Câu hỏi của Hoai Bao Tran

Question

tìm Min của:

$\sqrt{\dfrac{x^3}{x^3+8y^3}}+\sqrt{\dfrac{4y^3}{y^3+\left(x+y\right)^3}}$  với x,y >0

Lightning Farron · Accepted Answer

$T=\sqrt{\dfrac{x^3}{x^3+8y^3}}+\sqrt{\dfrac{4y^3}{y^3+\left(x+y\right)^3}}$

$=\dfrac{x^2}{\sqrt{x\left(x^3+8y^3\right)}}+\dfrac{2y^2}{\sqrt{y\left(y^3+\left(x+y\right)^3\right)}}$

$=\dfrac{x^2}{\sqrt{\left(x^2+2xy\right)\left(x^2-2xy+4y^2\right)}}+\dfrac{2y^2}{\sqrt{\left(xy+2y^2\right)\left(x^2+xy+y^2\right)}}$

$\ge\dfrac{2x^2}{2x^2+4y^2}+\dfrac{4y^2}{2y^2+\left(x+y\right)^2}$$\ge\dfrac{2x^2}{2x^2+4y^2}+\dfrac{4y^2}{4y^2+2x^2}$

$\ge\dfrac{2x^2+4y^2}{2x^2+4y^2}=1$Câu trả lời: $T=\sqrt{\dfrac{x^3}{x^3+8y^3}}+\sqrt{\dfrac{4y^3}{y^3+\left(x+y\right)^3}}$

$=\dfrac{x^2}{\sqrt{x\left(x^3+8y^3\right)}}+\dfrac{2y^2}{\sqrt{y\left(y^3+\left(x+y\right)^3\right)}}$

$=\dfrac{x^2}{\sqrt{\left(x^2+2xy\right)\left(x^2-2xy+4y^2\right)}}+\dfrac{2y^2}{\sqrt{\left(xy+2y^2\right)\left(x^2+xy+y^2\right)}}$

$\ge\dfrac{2x^2}{2x^2+4y^2}+\dfrac{4y^2}{2y^2+\left(x+y\right)^2}$$\ge\dfrac{2x^2}{2x^2+4y^2}+\dfrac{4y^2}{4y^2+2x^2}$

$\ge\dfrac{2x^2+4y^2}{2x^2+4y^2}=1$

Violympic toán 9