Matematické Fórum

stuart clark · 05. 11. 2011 16:15

If $a\;,b>0$ and $a^{13}+b^{13}=2$

Then prove that $\frac{5a^2}{b}+\frac{3b^3}{a^2}\geq 8$

laszky · 11. 03. 2018 01:05 — Editoval laszky (11. 03. 2018 01:21)

I have tried several AG-inequalities, but finally Lagrangian multipliers lead to the result.

Let us define a function $f(a,b)=\frac{5a^2}{b}+\frac{3b^3}{a^2}$ and a constraint function $g(a,b)=a^{13}+b^{13}-2$ . Then the function $f$ attains its extremal values in $a,b$ satisfying

$\nabla f(a,b) + \lambda\nabla g(a,b) =0$

This is a system of two equations

$10\frac{a}{b} - 6\frac{b^3}{a^3} + 13\lambda a^{12} = 0$

$-5\frac{a^2}{b^2} + 9\frac{b^2}{a^2} + 13\lambda b^{12} = 0$

$0=b^{12}.(EQ1) - a^{12}.(EQ2) = b^{12}\left(10\frac{a}{b} - 6\frac{b^3}{a^3}\right) - a^{12}\left(-5\frac{a^2}{b^2} + 9\frac{b^2}{a^2}\right)$

Hence, denoting $x=b/a$ we solve the polynomial equation

$p(x):=6x^{17}-10x^{13}+9x^4-5=0$

Obviously $x=1$ is the solution and if we show, that $p'(x)=(102x^{13}-130x^{9}+36)x^3$ is positive in $\mathbb{R}^{+}$ , then p is increasing in $\mathbb{R}^{+}$ and $x=1$ is the only solution.

We apply the Young inequality $uv\leq\frac{u^p}{p}+\frac{u^q}{q}$ with $p+q=pq$ . Here we choose

$p=\frac{13}{9}, \, q=\frac{13}{4}, \, u = \left(\frac{13}{9}\,102\right)^{\frac{9}{13}}x^9 \;\; \mathrm{and} \;\; v=130\left(\frac{13}{9}\,102\right)^{-\frac{9}{13}}$

Then

$130x^{9} = uv \leq \frac{u^p}{p}+\frac{u^q}{q} = 102x^{13} + \frac{4}{13}\,130^{\frac{13}{4}}\left(\frac{13}{9}\,102\right)^{-\frac{9}{4}} = 102x^{13} + 30.1825... < 102x^{13} + 36$ .

Thus, $x=b/a=1$ is the only solution and since from $a^{13}+a^{13}=2$ it follows $a=b=1$ we obtain $f(a,b)\geq f(1,1) = 5+3 = 8$ .

Matematické Fórum

#1 05. 11. 2011 16:15

Inequality

#2 11. 03. 2018 01:05 — Editoval laszky (11. 03. 2018 01:21)

Re: Inequality

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