Question

a,b là 2 số thực. x,y là 2 số thực dương

CMR \(\dfrac{a^2}{x}\)+\(\dfrac{b^2}{y}\)>\(\dfrac{\left(a+b\right)^2}{x+y}\)

Answer 1

Áp dụng bđt bunhia copski ta có

\(\left(a+b\right)^2=\left(\dfrac{a\sqrt{x}}{\sqrt{x}}+\dfrac{b\sqrt{y}}{\sqrt{y}}\right)^2\le\left[\left(\dfrac{a}{\sqrt{x}}\right)^2+\left(\dfrac{b}{\sqrt{y}}\right)^2\right]\left[\left(\sqrt{x}\right)^2+\left(\sqrt{y}\right)^2\right]=\left(\dfrac{a^2}{x}+\dfrac{b^2}{y}\right)\left(x+y\right)\Leftrightarrow\dfrac{a^2}{x}+\dfrac{b^2}{y}\ge\dfrac{\left(a+b\right)^2}{x+y}\)