Question

chứng minh rằng với mọi x,y lớn hơn 0 thì:

\(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{x^2+y^2}>=\frac{10}{\left(x+y\right)^2}\)

Answer 1

Áp dụng BĐT Cauchy, ta có:

 \(\frac{1}{x^2}+\frac{1}{y^2}\ge2\sqrt{\frac{1}{x^2}.\frac{1}{y^2}}=\frac{2}{xy}\)

\(\Rightarrow VT\ge\frac{2}{xy}+\frac{1}{x^2+y^2}\)

\(\Leftrightarrow VT\ge\left(\frac{1}{2xy}+\frac{1}{x^2+y^2}\right)+\frac{3}{2xy}\)

\(\Rightarrow VT\ge\frac{4}{\left(x+y\right)^2}+\frac{3}{\frac{\left(x+y\right)^2}{2}}\)

\(\Leftrightarrow VT\ge\frac{4}{\left(x+y\right)^2}+\frac{6}{\left(x+y\right)^2}=\frac{10}{\left(x+y\right)^2}\)

Dấu = xảy ra khi \(x=y>0\)

Vậy \(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{x^2+y^2}\ge\frac{10}{\left(x+y\right)^2}\) với \(\forall x;y>0\)