Question

Cho 0≤x,y≤\(\frac{\sqrt{2}}{2}\). Tìm GTLN của P=\(\frac{x}{1+y^2}\)+\(\frac{y}{1+x^2}\)

Answer 1

Lời giải:

Vì $0\leq x,y\leq \frac{\sqrt{2}}{2}$ nên:

$(2x^2-1)(2y^2-1)\geq 0\Leftrightarrow x^2y^2\geq \frac{2x^2+2y^2-1}{4}$

$P=\frac{x^3+x+y^3+y}{(x^2+1)(y^2+1)}=\frac{x^3+x+y^3+y}{x^2y^2+x^2+y^2+1}\leq \frac{x^3+y^3+x+y}{\frac{2x^2+2y^2-1}{4}+x^2+y^2+1}=\frac{4}{3}.\frac{x^3+y^3+x+y}{2x^2+2y^2+1}$

Giờ ta sẽ chứng minh:

$\frac{x^3+y^3+x+y}{2x^2+2y^2+1}\leq \frac{\sqrt{2}}{2}(*)$

Thật vậy:

\(x^3+x-\sqrt{2}x^2-\frac{1}{2\sqrt{2}}=(x-\frac{1}{\sqrt{2}})[x^2-\frac{1}{\sqrt{2}}x+\frac{1}{2}]=(x-\frac{1}{\sqrt{2}})[(x-\frac{1}{2\sqrt{2}})+\frac{3}{8}]^2\leq 0\) với mọi $x\leq \frac{1}{2\sqrt{2}}$

$y^3+y-\sqrt{2}y^2-\frac{1}{2\sqrt{2}}\leq 0$

$\Rightarrow x^3+y^3+x+y\leq \frac{1}{\sqrt{2}}(2x^2+2y^2+1)$

$\Rightarrow (*)$ đúng

$\Rightarrow P\leq \frac{4}{3}.\frac{1}{\sqrt{2}}=\frac{2\sqrt{2}}{3}$

Vậy...........