Câu hỏi của Phạm Tuấn Long

Question

B1: Cho x,y là các số dương thỏa mãn x+y$\le$1 . Tìm giá trị nhỏ nhất của biểu thức sau :

A=$\frac{1}{x^3+3xy^2}+\frac{1}{y^3+3x^2y}$

help me !!!

Nguyệt Dạ · Accepted Answer

_Solution:

Prove with Cauchy-Schwarz inequality engel form, we have:

$A=\frac{1}{x^3+3xy^2}+\frac{1}{y^3+3x^2y}\ge\frac{4}{x^3+y^3+3xy^2+3x^2y}$

$A\ge\frac{4}{\left(x+y\right)^3}$

Other way: $x+y\le1\Rightarrow\left(x+y\right)^3\le1\Rightarrow\frac{1}{\left(x+y\right)^3}\ge1$

$\Rightarrow A\ge4$ (proof)

We have ''='' $\Leftrightarrow x=y=\frac{1}{2}$.Câu trả lời: _Solution:

Prove with Cauchy-Schwarz inequality engel form, we have:

$A=\frac{1}{x^3+3xy^2}+\frac{1}{y^3+3x^2y}\ge\frac{4}{x^3+y^3+3xy^2+3x^2y}$

$A\ge\frac{4}{\left(x+y\right)^3}$

Other way: $x+y\le1\Rightarrow\left(x+y\right)^3\le1\Rightarrow\frac{1}{\left(x+y\right)^3}\ge1$

$\Rightarrow A\ge4$ (proof)

We have ''='' $\Leftrightarrow x=y=\frac{1}{2}$.