Mathematics

IneqMath

IneqMath: Olympiad-level inequality problems reformulated into bound estimation and relation prediction subtasks. Binary response = the final-answer-judge correctness (deterministically reconstructed from the released per-problem responses and validated to match the released scores.json aggregates). Upstream releases full per-problem output for one worked-example model (gpt-4o-mini); the 50+ leaderboard models are aggregate-only.

100items

1subjects

100%observed

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Response matrix

Every model, scored item by item.

Each row is an AI model and each column an item, ordered so the strongest models and easiest items gather toward one corner. 1 subjects × 100 items, 100% of cells evaluated.

IneqMath response matrix: AI models (rows) against items (columns) — Correct (1)Incorrect (0)Unobserved
Scale: 1 = correct · 0 = incorrect

Sample items

What the questions look like — and how subjects answer.

A spread of items across the difficulty range, each shown with a few subjects' actual answers.

Item 10% solve rate

Let $x, y, z \in \mathbb{R}^{+}$ such that $\sqrt{x} + \sqrt{y} + \sqrt{z} = 1$ . Consider the following inequality:

(1-x)^2(1-y)^2(1-z)^2 \quad () \quad 2^{15} x y z (x+y)(y+z)(z+x).

Determine the correct inequality relation to fill in the blank.

Options:

A.
$\leq$
B.
$\geq$
C.
$=$
D.
$<$
E.
$>$
F.
None of the above

How subjects answered

gpt-4o-mini incorrect
To solve the problem, we start with the given condition:

[ \sqrt{x} + \sqrt{y} + \sqrt{z} = 1 ]

We need to analyze the inequality:

[ (1-x)^2(1-y)^2(1-z)^2 \quad () \quad 2^{15} x y z (x+y)(y+z)(z+x) ]

First, we express (x), (y), and (z) in terms of (a), (b), and (c) where (a = \sqrt{x}), (b = \sqrt{y}), and (c = \sqrt{z}). Thus, we have:

[ x = a^2, \quad y = b^2, \quad z = c^2 ]

The condition becomes:

[ a + b + c = 1 ]

Next, we can express (1-x), (1-y), and (1-z):

[ 1-x = 1-a^2, \quad 1-y = 1-b^2, \quad 1-z = 1-c^2 ]

Now, we can rewrite the left-hand side of the inequality:

[ (1-x)^2(1-y)^2(1-z)^2 = (1-a^2)^2(1-b^2)^2(1-c^2)^2 ]

Next, we …

Item 20% solve rateanswer: $C = 64$

Let $a, b, c, d \in \mathbb{R}^{+}$ . Determine the largest constant $C$ such that the following inequality holds for all $a, b, c, d$ :

\frac{1}{a}+\frac{1}{b}+\frac{4}{c}+\frac{16}{d} \geq \frac{C}{a+b+c+d}

How subjects answered

gpt-4o-mini incorrect
To solve the problem, we need to find the largest constant ( C ) such that the following inequality holds for all positive real numbers ( a, b, c, d ):

[ \frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d} \geq \frac{C}{a+b+c+d}. ]

Step 1: Apply the Cauchy-Schwarz Inequality

We can apply the Cauchy-Schwarz inequality in the following form:

[ \left( \frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d} \right) (a + b + c + d) \geq (1 + 1 + 4 + 16)^2. ]

Calculating the right-hand side:

[ 1 + 1 + 4 + 16 = 22, ] thus,

[ (1 + 1 + 4 + 16)^2 = 22^2 = 484. ]

Step 2: Rearranging the Inequality

From the Cauchy-Schwarz inequality, we have:

[ \frac{1}{a} + \frac{1}{b} …

Item 30% solve rate

Let $a, b, c > 0$ be positive real numbers such that $a + b + c = 3$ . Consider the following inequality:

a^2 + 4b^2 + 4b + 8c \quad () \quad 7 + 2\sqrt{2(b^2 + c^2)} + 4\sqrt{bc}.

Determine the correct inequality relation to fill in the blank.

Options:

A.
$\leq$
B.
$\geq$
C.
$=$
D.
$<$
E.
$>$
F.
None of the above

How subjects answered

gpt-4o-mini incorrect
To solve the inequality ( a^2 + 4b^2 + 4b + 8c \quad () \quad 7 + 2\sqrt{2(b^2 + c^2)} + 4\sqrt{bc} ) under the condition ( a + b + c = 3 ), we will analyze both sides of the inequality.

Step 1: Express ( a ) in terms of ( b ) and ( c )

Since ( a + b + c = 3 ), we can express ( a ) as: [ a = 3 - b - c ]

Step 2: Substitute ( a ) into the left-hand side

Substituting ( a ) into the left-hand side of the inequality gives: [ (3 - b - c)^2 + 4b^2 + 4b + 8c ]

Expanding ( (3 - b - c)^2 ): [ (3 - b - c)^2 = 9 - 6b - 6c + b^2 + 2bc + c^2 ]

Thus, the left-hand side becomes: [ 9 - 6b - 6c + b^2 + 2bc + c^2 + 4b^2 + 4b + 8c ] Combining like terms: [ = 9 + 5b^2 …

Item 40% solve rate

Let $a, b, c$ be positive numbers. Consider the following inequality:

\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \quad () \quad \frac{5 a}{2 a^2+b^2+c^2}+\frac{5 b}{a^2+2 b^2+c^2}+\frac{5 c}{a^2+b^2+2 c^2}.

Determine the correct inequality relation to fill in the blank.

Options:

A.
$\leq$
B.
$\geq$
C.
$=$
D.
$<$
E.
$>$
F.
None of the above

How subjects answered

gpt-4o-mini incorrect
To solve the inequality

[ \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \quad () \quad \frac{5 a}{2 a^2+b^2+c^2}+\frac{5 b}{a^2+2 b^2+c^2}+\frac{5 c}{a^2+b^2+2 c^2}, ]

we will analyze both sides of the inequality.

Step 1: Analyze the left-hand side

The left-hand side is

[ \frac{1}{a} + \frac{1}{b} + \frac{1}{c}. ]

Step 2: Analyze the right-hand side

The right-hand side is

[ \frac{5 a}{2 a^2 + b^2 + c^2} + \frac{5 b}{a^2 + 2 b^2 + c^2} + \frac{5 c}{a^2 + b^2 + 2 c^2}. ]

Step 3: Apply Cauchy-Schwarz Inequality

We can apply the Cauchy-Schwarz inequality to the right-hand side. By Cauchy-Schwarz, we have:

[ \left( \frac{5 a}{2 a^2 + b^2 + c^2} + \frac{5 b}{a^2 + 2 b^2 + …

Item 50% solve rate

Let $a, b, c \in \mathbb{R}^{+}$ be positive real numbers such that $abc = 1$ . Consider the following expressions:

a^3 + b^3 + 2c^3 \quad () \quad a^2 + b^2 + c^2.

Determine the correct inequality relation to fill in the blank.

Options:

A.
$\leq$
B.
$\geq$
C.
$=$
D.
$<$
E.
$>$
F.
None of the above

How subjects answered

gpt-4o-mini incorrect
To solve the problem, we start with the given condition that ( abc = 1 ) for positive real numbers ( a, b, c ). We need to analyze the inequality:

[ a^3 + b^3 + 2c^3 \quad () \quad a^2 + b^2 + c^2. ]

Step 1: Use the identity for cubes

We can use the identity for the sum of cubes:

[ a^3 + b^3 = (a+b)(a^2 - ab + b^2). ]

Step 2: Apply the AM-GM inequality

Since ( abc = 1 ), we can express ( c ) in terms of ( a ) and ( b ):

[ c = \frac{1}{ab}. ]

Now substituting ( c ) into the inequality, we have:

[ a^3 + b^3 + 2\left(\frac{1}{ab}\right)^3 \quad () \quad a^2 + b^2 + \left(\frac{1}{ab}\right)^2. ]

Step 3: Simplify the expressions

The left-hand side …

Item 650% solve rateanswer: $C = 1$

Given non-negative real numbers $a, b, c$ satisfying $ab + bc + ca = 1$ , find the largest constant $C$ such that the following inequality holds for all $a, b, c$ :

\frac{c}{\sqrt{a+b+3c}} + \frac{a}{\sqrt{c+b+3a}} + \frac{b}{\sqrt{a+c+3b}} \geq C.

How subjects answered

gpt-4o-mini incorrect
To solve the problem, we need to find the largest constant ( C ) such that the following inequality holds for all non-negative real numbers ( a, b, c ) satisfying ( ab + bc + ca = 1 ):

[ \frac{c}{\sqrt{a+b+3c}} + \frac{a}{\sqrt{c+b+3a}} + \frac{b}{\sqrt{a+c+3b}} \geq C. ]

Step 1: Analyze the expression

We start by analyzing the left-hand side of the inequality. We can use the Cauchy-Schwarz inequality to help us bound the terms.

Step 2: Apply Cauchy-Schwarz Inequality

Using the Cauchy-Schwarz inequality, we have:

[ \left( \frac{c}{\sqrt{a+b+3c}} + \frac{a}{\sqrt{c+b+3a}} + \frac{b}{\sqrt{a+c+3b}} \right)^2 \leq (c+a+b) \left( \frac{c^2}{a+b+3c} + \frac{a^2}{c+b+3a} + …

Item 750% solve rateanswer: $C = 2$

Let $a, b, c > 0$ and $k \in \mathbb{N}^{+}$ . Determine the largest constant $C$ such that the following inequality holds for all $a, b, c$ :

\left(a^2+\frac{2(k+1)^2}{b+k}\right)\left(b^2+\frac{2(k+1)^2}{c+k}\right)\left(c^2+\frac{2(k+1)^2}{a+k}\right) \geq (Ck+3)^3.

How subjects answered

gpt-4o-mini incorrect
To solve the problem, we need to analyze the inequality given:

[ \left(a^2+\frac{2(k+1)^2}{b+k}\right)\left(b^2+\frac{2(k+1)^2}{c+k}\right)\left(c^2+\frac{2(k+1)^2}{a+k}\right) \geq (Ck+3)^3. ]

Step 1: Simplifying the Terms

We start by rewriting the terms inside the product. For each term, we can apply the AM-GM inequality:

[ a^2 + \frac{2(k+1)^2}{b+k} \geq 3\sqrt[3]{a^2 \cdot \frac{2(k+1)^2}{b+k} \cdot \frac{2(k+1)^2}{b+k}}. ]

This gives us a lower bound for each factor in the product.

Step 2: Applying AM-GM

Using AM-GM on each term, we have:

[ a^2 + \frac{2(k+1)^2}{b+k} \geq 3\sqrt[3]{a^2 \cdot \left(\frac{2(k+1)^2}{b+k}\right)^2}. ]

Thus, we can …

Item 850% solve rateanswer: $C = 27$

Given positive real numbers $a, b, c$ satisfying $ab + bc + ca = 3$ , find the largest constant $C$ such that the following inequality holds for all $a, b, c$ :

\frac{1}{\sqrt{a+b}} + \frac{1}{\sqrt{c+b}} + \frac{1}{\sqrt{a+c}} \geq \sqrt{\frac{C}{a+b+c+3}}.

How subjects answered

gpt-4o-mini incorrect
To solve the problem, we need to find the largest constant ( C ) such that the following inequality holds for all positive real numbers ( a, b, c ) satisfying ( ab + bc + ca = 3 ):

[ \frac{1}{\sqrt{a+b}} + \frac{1}{\sqrt{b+c}} + \frac{1}{\sqrt{c+a}} \geq \sqrt{\frac{C}{a+b+c+3}}. ]

Step 1: Analyze the left-hand side

Using the Cauchy-Schwarz inequality, we can estimate the left-hand side:

[ \left( \frac{1}{\sqrt{a+b}} + \frac{1}{\sqrt{b+c}} + \frac{1}{\sqrt{c+a}} \right)^2 \leq (1 + 1 + 1) \left( \frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{c+a} \right). ]

This simplifies to:

[ \frac{1}{\sqrt{a+b}} + \frac{1}{\sqrt{b+c}} + \frac{1}{\sqrt{c+a}} \leq \sqrt{3} …

Item 9100% solve rate

Let $a, b, c > 0$ be positive real numbers. Consider the following inequality:

\frac{a^2}{a^2+bc} + \frac{b^2}{b^2+ca} + \frac{c^2}{c^2+ba} \quad () \quad \frac{(a+b+c)^2}{2(ab+bc+ca)}.

Determine the correct inequality relation to fill in the blank.

Options:

A.
$\leq$
B.
$\geq$
C.
$=$
D.
$<$
E.
$>$
F.
None of the above

How subjects answered

gpt-4o-mini correct
To solve the inequality

[ \frac{a^2}{a^2+bc} + \frac{b^2}{b^2+ca} + \frac{c^2}{c^2+ab} \quad () \quad \frac{(a+b+c)^2}{2(ab+bc+ca)}, ]

we will analyze both sides of the inequality.

Step 1: Analyze the Left Side

We start with the left-hand side:

[ L = \frac{a^2}{a^2+bc} + \frac{b^2}{b^2+ca} + \frac{c^2}{c^2+ab}. ]

Using the Cauchy-Schwarz inequality, we can derive a useful bound for (L):

[ L = \sum_{\text{cyc}} \frac{a^2}{a^2 + bc} \geq \frac{(a+b+c)^2}{\sum_{\text{cyc}} (a^2 + bc)}. ]

Step 2: Analyze the Right Side

Now, we analyze the right-hand side:

[ R = \frac{(a+b+c)^2}{2(ab+bc+ca)}. ]

Step 3: Compare (L) and (R)

From the Cauchy-Schwarz inequality, …

Item 10100% solve rateanswer: $C = 7$

Let $a, b, c \geq 0$ with $ab+bc+ca > 0$ . Determine the largest constant $C$ such that the following inequality holds for all $a, b, c$ :

\frac{a}{\sqrt{b+c+Ca}} + \frac{b}{\sqrt{a+c+Cb}} + \frac{c}{\sqrt{b+a+Cc}} \geq \sqrt{\frac{ab+bc+ca}{a+b+c}}

How subjects answered

gpt-4o-mini correct
To solve the problem, we need to find the largest constant ( C ) such that the following inequality holds for all non-negative ( a, b, c ) with ( ab + bc + ca > 0 ):

[ \frac{a}{\sqrt{b+c+Ca}} + \frac{b}{\sqrt{a+c+Cb}} + \frac{c}{\sqrt{b+a+Cc}} \geq \sqrt{\frac{ab+bc+ca}{a+b+c}}. ]

Step 1: Analyze the left-hand side

We start by analyzing the left-hand side of the inequality. We can apply the Cauchy-Schwarz inequality:

[ \left( \frac{a}{\sqrt{b+c+Ca}} + \frac{b}{\sqrt{a+c+Cb}} + \frac{c}{\sqrt{b+a+Cc}} \right)^2 \leq (a+b+c) \left( \frac{a^2}{b+c+Ca} + \frac{b^2}{a+c+Cb} + \frac{c^2}{b+a+Cc} \right). ]

Step 2: Simplify the right-hand side

We need to relate the …

Item 11100% solve rate

Let $a, b, c \geq 0$ be non-negative real numbers such that $a + b + c = 1$ . Consider the following inequality:

7(ab + bc + ca) \quad () \quad 3 + 9abc.

Determine the correct inequality relation to fill in the blank.

Options:

A.
$\leq$
B.
$\geq$
C.
$=$
D.
$<$
E.
$>$
F.
None of the above

How subjects answered

gpt-4o-mini correct
To solve the inequality ( 7(ab + bc + ca) \quad () \quad 3 + 9abc ) under the condition ( a + b + c = 1 ) where ( a, b, c \geq 0 ), we will analyze both sides of the inequality.

Express ( ab + bc + ca ): Using the identity ( (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + ac + bc) ), we can express ( ab + ac + bc ) in terms of ( a, b, c ): [ ab + ac + bc = \frac{(a + b + c)^2 - (a^2 + b^2 + c^2)}{2} = \frac{1 - (a^2 + b^2 + c^2)}{2}. ]

Express ( abc ): The product ( abc ) is maximized when ( a, b, c ) are as equal as possible. By the AM-GM inequality, we have: [ abc \leq \left(\frac{a + b + c}{3}\right)^3 = \left(\frac{1}{3}\right)^3 = …

Item 12100% solve rate

Let $a, b, c > 0$ be positive real numbers such that $ab + bc + ca = 1$ . Consider the following inequality:

a b c\left(a+\sqrt{a^2+1}\right)\left(b+\sqrt{b^2+1}\right)\left(c+\sqrt{c^2+1}\right) \quad () \quad 2.

Determine the correct inequality relation to fill in the blank.

Options:

A.
$\leq$
B.
$\geq$
C.
$=$
D.
$<$
E.
$>$
F.
None of the above

How subjects answered

gpt-4o-mini correct
To solve the problem, we start with the given condition ( ab + bc + ca = 1 ) and need to analyze the expression

[ abc\left(a+\sqrt{a^2+1}\right)\left(b+\sqrt{b^2+1}\right)\left(c+\sqrt{c^2+1}\right). ]

First, we can simplify the terms ( a + \sqrt{a^2 + 1} ). Notice that:

[ a + \sqrt{a^2 + 1} = \sqrt{a^2 + 1} + a = \sqrt{a^2 + 1} + a \geq \sqrt{2a^2} = a\sqrt{2}, ]

since ( \sqrt{a^2 + 1} \geq \sqrt{a^2} = a ).

Thus, we can write:

[ a + \sqrt{a^2 + 1} \geq a\sqrt{2}. ]

Applying this to each of the three terms, we have:

[ \left(a + \sqrt{a^2 + 1}\right)\left(b + \sqrt{b^2 + 1}\right)\left(c + \sqrt{c^2 + 1}\right) \geq (a\sqrt{2})(b\sqrt{2})(c\sqrt{2}) = …

Subjects

The models, agents, and reward models evaluated.

1 subjects, ranked by mean response (accuracy) across this benchmark's items.

1gpt-4o-mini0.545

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