tìm max

Question

Cho các số thực dương

$x,y,z$ thỏa mãn

$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{10}{x+y+z}$ .Tính giá trị lớn nhất của

$P=\frac{3}{xy+yz+zx}-\frac{4}{x^{3}+y^{3}+z^{3}}$

Confusion · Answer 1 · 6/15/2016 1:27:25 PM

No decrease of generality, assume

$x\leq y\leq z$

Set

$t=x+y+z$

From the assumptions, applied inequality Cauchy, we have:

$7=\frac{x+y}{z}+z(\frac{1}{x}+\frac{1}{y})+\frac{x}{y}+\frac{y}{x}\geq \frac{2\sqrt{xy}}{z}+\frac{2z}{\sqrt{xy}}+2$

So that:

$2\sqrt{xy}\geq z\rightarrow x+y\geq z$

$\Rightarrow (x+y-z)(y+z-x)(z+x-y)\geq 0$

$\Leftrightarrow x^3+y^2+z^3\leq \sum xy(x+y)-2xyz$

Assumptions rewritten as:

$(x+y+z)(xy+yz+zx)=10xyz$

$\rightarrow$ We have:

$\sum xy(x+y)=7xyz$

and:

$\prod (x+y)=9xyz$

So that:

$x^3+y^3+z^3\leq \sum xy(x+y)-2xyz=5xyz$

We always have:

$x^3+y^3+z^3+3\prod (x+y)=(x+y+z)^3$

$\rightarrow x^3+y^3+z^3\leq \frac{5}{9}\prod (x+y)\leq \frac{5}{32}t^3$

and:

$xy+yz+zx=\frac{10xyz}{x+y+z}\geq \frac{5}{16}.\frac{x^3+y^3+z^3+3\prod (x+y)}{t}=\frac{5}{16}t^2$

So:

$P\leq \frac{48}{5t^2}-\frac{128}{5t^3}=f(t)$

We have:

$f'(t)=\frac{96(4-t)}{5t^4}$

Tabulation of change of

$f (t)$ we get

$f(t)\leq f(4)=\frac{1}{5}$

$\rightarrow P\leq \frac{1}{5}$ at

$x=y=1;z=2./$

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