Question

Bài 1:Cho a, b là 2 số bất kì và x, y là 2 số dương.CM

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

Answer 1

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

\(\Rightarrow\dfrac{a^2}{x}+\dfrac{b^2}{y}-\dfrac{\left(a+b\right)^2}{x+y}\ge0\)

\(\Rightarrow\dfrac{a^2}{x}+\dfrac{b^2}{y}-\dfrac{a^2}{x+y}-\dfrac{b^2}{x+y}-\dfrac{2ab}{x+y}\)

\(\Rightarrow\dfrac{a^2x+a^2y-a^2x}{x\left(x+y\right)}+\dfrac{b^2x+b^2y-b^2y}{y\left(x+y\right)}-\dfrac{2ab}{x+y}\)\(\Rightarrow\dfrac{a^2y^2+b^2x^2-2abxy}{xy\left(x+y\right)}=\dfrac{\left(ay-bx\right)^2}{x^2y+xy^2}\ge0\)luôn đúng \(\Rightarrowđpcm\)

Answer 2

I do not know