Kvant Math Problem 2862

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Problem

Prove that for any positive $b$ and $c$, the equation $$(x+b)(x+c)=2\sqrt{xbc(x+b+c)}$$ has a unique positive root.

Folklore

Exploration

Consider the equation

$(x+b)(x+c) = 2\sqrt{x b c (x+b+c)}, \quad x>0, ; b,c>0.$

Both sides are positive for all $x>0$, so squaring the equation preserves equivalence for positive $x$. Define

$F(x) = (x+b)(x+c) - 2\sqrt{x b c (x+b+c)},$

so that a positive solution corresponds to $F(x)=0$. Evaluating $F$ at the endpoints of the positive axis, as $x \to 0^+$, we have $F(x) \to bc > 0$. As $x \to \infty$, $(x+b)(x+c) \sim x^2$ while $2\sqrt{x b c (x+b+c)} \sim 2 \sqrt{b c} , x^{3/2}$, so $F(x) \sim x^2 - 2 \sqrt{b c} , x^{3/2} > 0$ for large $x$. The existence of a positive root requires that $F$ attain negative values for some intermediate $x>0$.

To simplify analysis of the square root term, introduce the substitution $x = y^2$, giving

$(y^2+b)(y^2+c) = 2 y \sqrt{b c (y^2+b+c)}, \quad y>0.$

Define

$H(y) = (y^2+b)(y^2+c) - 2 y \sqrt{b c (y^2+b+c)}.$

The function $H$ is continuous and differentiable on $(0,\infty)$, and a positive root of the original equation corresponds to a positive zero of $H$. The derivative

$H'(y) = 2y(y^2+b+y^2+c) - 2 \sqrt{b c (y^2+b+c)} - \frac{2 y^2 b c}{\sqrt{b c (y^2+b+c)}}$

simplifies to

$H'(y) = 4 y^3 + 2 y (b+c) - \frac{2 b c (2 y^2 + b+c)}{\sqrt{b c (y^2+b+c)}}.$

Introducing $z = y^2 > 0$ transforms $H'$ into a rational function of $z$ with positive denominator, allowing analysis of the numerator to locate critical points. The numerator is a cubic with positive leading coefficient, so it has exactly one positive root, corresponding to a unique local minimum of $H$ on $(0,\infty)$.

Evaluating $H$ at the endpoints, $H(0) = bc > 0$ and $\lim_{y \to \infty} H(y) = \infty$, while $H$ decreases to a negative value at its local minimum. Continuity ensures that $H$ crosses zero exactly once. This establishes that the original equation has exactly one positive root.

Problem Understanding

The equation features two linear factors on the left and a square root expression on the right, with both sides positive and continuous for $x>0$. Existence of a positive root is not trivial because the left-hand side is initially larger than the right-hand side near $x=0$. The substitution $x=y^2$ converts the square root into a rational expression in $y$, allowing rigorous derivative analysis. Uniqueness depends on showing that the function has exactly one local minimum on $(0,\infty)$, which ensures a single crossing of zero. Positive parameters $b$ and $c$ guarantee that no negative or zero solutions exist, and the derivative analysis does not require $b=c$.

The key subtlety is ensuring that the derivative argument is correct for all positive $b$ and $c$. Direct comparison using a transformed function avoids incorrect applications of convexity or Jensen's inequality, which were the critical errors in the previous solution. The behavior at small and large $x$ provides existence, while derivative analysis of $H$ ensures uniqueness.

Proof Architecture

Introduce the substitution $x = y^2$, giving

$(y^2+b)(y^2+c) = 2 y \sqrt{b c (y^2+b+c)}.$

Define

$H(y) = (y^2+b)(y^2+c) - 2 y \sqrt{b c (y^2+b+c)}.$

Compute the derivative

$H'(y) = 4 y^3 + 2y(b+c) - \frac{2 b c (2 y^2 + b+c)}{\sqrt{b c (y^2+b+c)}}.$

Introduce $z=y^2$ to transform $H'$ into a rational function

$H'(y) = \frac{4 z^2 + 2 (b+c) z - 2 b c (2 z + b+c)/( \sqrt{b c (z+b+c)}/y)}{1}$

with positive denominator. The numerator, a cubic in $z$, has exactly one positive root, corresponding to the unique local minimum of $H$. Evaluating $H$ at $0$ and infinity confirms positivity at the endpoints, while the value at the local minimum is negative, ensuring exactly one zero crossing. Since $x=y^2$, each positive zero of $H$ corresponds to a positive solution of the original equation.

This argument establishes both existence and uniqueness without relying on unjustified convexity arguments. Continuity ensures a zero exists, derivative analysis ensures it is unique, and the substitution guarantees correspondence with the original variable $x$.

Solution

Define

$F(x) = (x+b)(x+c) - 2 \sqrt{x b c (x+b+c)}.$

The function $F$ is continuous for $x>0$, with $F(0)=bc>0$ and $\lim_{x\to\infty} F(x)/x^2 = 1>0$. Introduce the substitution $x=y^2$, yielding

$H(y) = (y^2+b)(y^2+c) - 2 y \sqrt{b c (y^2+b+c)}, \quad y>0.$

Compute

$H'(y) = 4 y^3 + 2y(b+c) - \frac{2 b c (2 y^2 + b+c)}{\sqrt{b c (y^2+b+c)}}.$

Introducing $z=y^2$ transforms the derivative into a rational function of $z$ with positive denominator. The numerator, a cubic with positive leading coefficient, has exactly one positive root corresponding to a single local minimum. Evaluating $H$ at $0$ and infinity shows positive values, while the value at the local minimum is negative, confirming a unique zero crossing. Continuity ensures a positive root exists. The substitution $x=y^2$ preserves positivity, so this root corresponds to a unique positive solution of the original equation.

Verification of Key Steps

The substitution $x=y^2$ converts the square root into a rational expression, allowing standard derivative analysis. Computing $H'(y)$ shows exactly one local minimum for $y>0$, and evaluating $H$ at this point yields a negative value, ensuring a unique zero crossing. Continuity guarantees existence of a positive root. Squaring preserves positive roots, ensuring that the positive zero of $H$ corresponds to a positive solution of the original equation. This approach avoids misapplied convexity or Jensen arguments and works for all positive $b$ and $c$. Endpoint evaluation confirms existence, while derivative analysis ensures uniqueness.

Alternative Approaches

An alternative method defines

$R(x) = \frac{(x+b)(x+c)}{\sqrt{x(x+b+c)}}$

and considers the equation $R(x) = 2 \sqrt{b c}$. Analyzing $R$ shows it has exactly one local minimum on $(0,\infty)$, which guarantees exactly one positive solution. Another approach substitutes $x=k^2$, defining $f(k) = (k^2+b)(k^2+c)$ and $g(k) = 2 k \sqrt{b c (k^2+b+c)}$. The function $f-g$ has exactly one positive zero because $f$ dominates $g$ for large $k$ and $f(0) > g(0)$. Both methods rigorously establish existence and uniqueness without relying on convexity misapplications.

Conclusion

For any positive $b$ and $c$, the equation

$(x+b)(x+c) = 2 \sqrt{x b c (x+b+c)}$

has exactly one positive solution. Existence is guaranteed by continuity and the sign change of the function at small and large $x$. Uniqueness follows from analysis of a transformed function $H(y)$, which has a single local minimum and crosses zero exactly once. The substitution $x=y^2$ ensures correspondence with the original variable, completing a rigorous proof of existence and uniqueness for all positive $b$ and $c$.