Kvant Math Problem 374

The expression contains two square roots that both include the factor $c$, suggesting a factorization by $\sqrt{c}$.

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Problem

M374. Let $a, b$ and $c$ be positive numbers, $a > c$ and $b > c$. Prove the inequality $\sqrt{c(a-c)} + \sqrt{c(b-c)} \le \sqrt{ab}$.

Exploration

The expression contains two square roots that both include the factor $c$, suggesting a factorization by $\sqrt{c}$. Writing

$\sqrt{c(a-c)} + \sqrt{c(b-c)} = \sqrt{c}\bigl(\sqrt{a-c} + \sqrt{b-c}\bigr)$

reduces the problem to comparing a sum of two square roots with $\sqrt{ab/c}$ after squaring the right-hand side.

Squaring is viable because all quantities are positive. After squaring, mixed terms of the form $\sqrt{(a-c)(b-c)}$ appear, which suggests completing a square after introducing $x = \sqrt{(a-c)(b-c)}$. The structure likely collapses into a perfect square.

Problem Understanding

This is a Type B problem: prove the inequality

$\sqrt{c(a-c)} + \sqrt{c(b-c)} \le \sqrt{ab},$

for positive $a,b,c$ with $a>c$ and $b>c$.

The key difficulty is handling the interaction between the two square roots after squaring the inequality, and ensuring the resulting algebraic expression can be reduced to a nonnegative form.

Proof Architecture

The proof begins by factoring out $\sqrt{c}$ from the left-hand side.

The main lemma is that after squaring both sides, the inequality is equivalent to

$(a-c)(b-c) + c^2 \ge 2c\sqrt{(a-c)(b-c)}.$

The second lemma is the substitution $x = \sqrt{(a-c)(b-c)}$, which transforms the inequality into a perfect square:

$x^2 - 2cx + c^2 \ge 0.$

The hardest step is verifying that the algebraic rearrangement after squaring introduces no extraneous asymmetry and preserves equivalence under positivity constraints.

Solution

Since $a>c$ and $b>c$, all expressions under square roots are positive. The given inequality is equivalent to

$\sqrt{c}\bigl(\sqrt{a-c} + \sqrt{b-c}\bigr) \le \sqrt{ab}.$

Squaring both sides yields

$c\bigl(\sqrt{a-c} + \sqrt{b-c}\bigr)^2 \le ab.$

Expanding the square gives

$c\bigl((a-c) + (b-c) + 2\sqrt{(a-c)(b-c)}\bigr) \le ab.$

Distributing $c$ leads to

$c(a+b-2c) + 2c\sqrt{(a-c)(b-c)} \le ab.$

Rearranging terms gives

$ab - c(a+b-2c) \ge 2c\sqrt{(a-c)(b-c)}.$

The left-hand side simplifies as

$ab - ca - cb + 2c^2 = (a-c)(b-c) + c^2.$

Thus the inequality becomes

$(a-c)(b-c) + c^2 \ge 2c\sqrt{(a-c)(b-c)}.$

Let $x = \sqrt{(a-c)(b-c)}$. Then $x>0$ and the inequality becomes

$x^2 + c^2 \ge 2cx.$

This is equivalent to

$(x-c)^2 \ge 0,$

which holds for all real $x$ and $c$. Therefore the original inequality holds.

This completes the proof. ∎

Verification of Key Steps

The first delicate step is squaring the inequality. Both sides are nonnegative because $c>0$ and $a>c$, $b>c$ ensure each square root is defined and positive, so squaring preserves equivalence.

The second step is the algebraic identity

$ab - ca - cb + 2c^2 = (a-c)(b-c) + c^2,$

which follows by direct expansion of $(a-c)(b-c)$ and comparison of coefficients.

The third step is the reduction to a perfect square. Substituting $x=\sqrt{(a-c)(b-c)}$ yields

$x^2 + c^2 - 2cx = (x-c)^2,$

which is obtained by expanding the square and matching terms exactly.

Alternative Approaches

A different approach uses the substitution $a = c + x^2$ and $b = c + y^2$, which transforms the inequality into

$c(x+y) \le \sqrt{(c+x^2)(c+y^2)}.$

Squaring again leads to a symmetric inequality in $x$ and $y$ that reduces to $(x-y)^2 \ge 0$ after cancellation. This method exposes symmetry earlier but introduces more algebraic substitution steps than the direct completion-of-square argument.