Cho m,n >0. CMR: $\frac{ a ²}{m}$ + $\frac{ b ²}{n}$ ≥ $\frac{(a+b) ²}{m+n}$ 31/07/2021 Bởi Ivy Cho m,n >0. CMR: $\frac{ a ²}{m}$ + $\frac{ b ²}{n}$ ≥ $\frac{(a+b) ²}{m+n}$
Giải thích các bước giải: Ta có: $(\dfrac{a^2}{m}+\dfrac{b^2}{n})(m+n)=a^2+b^2+\dfrac{a^2n}{m}+\dfrac{b^2m}{n}$ $\to (\dfrac{a^2}{m}+\dfrac{b^2}{n})(m+n)\ge a^2+b^2+2\sqrt{\dfrac{a^2n}{m}\cdot\dfrac{b^2m}{n}}$ $\to (\dfrac{a^2}{m}+\dfrac{b^2}{n})(m+n)\ge a^2+b^2+2ab$ $\to (\dfrac{a^2}{m}+\dfrac{b^2}{n})(m+n)\ge (a+b)^2$ $\to \dfrac{a^2}{m}+\dfrac{b^2}{n}\ge \dfrac{(a+b)^2}{m+n}$ Bình luận
$\dfrac{a^2}{m} + \dfrac{b^2}{n} \geq \dfrac{(a+b)^2}{m+n}$ $\Leftrightarrow \dfrac{a^2n + b^2m}{mn} \geq \dfrac{a^2 + b^2 + 2ab}{m+ n}$ $\Leftrightarrow (a^2n + b^2m)(m+n) \geq (a^2 +b^2 +2ab)mn$ $\Leftrightarrow a^2mn + a^2n^2 + b^2m^2 + b^2mn \geq a^2mn + b^2mn + 2abmn$ $\Leftrightarrow a^2n^2 – 2abmn + b^2m^2 \geq 0$ $\Leftrightarrow (an – bm)^2 \geq 0$ (luôn đúng) Vậy $\dfrac{a^2}{m} + \dfrac{b^2}{n} \geq \dfrac{(a+b)^2}{m+n}$ Bình luận
Giải thích các bước giải:
Ta có:
$(\dfrac{a^2}{m}+\dfrac{b^2}{n})(m+n)=a^2+b^2+\dfrac{a^2n}{m}+\dfrac{b^2m}{n}$
$\to (\dfrac{a^2}{m}+\dfrac{b^2}{n})(m+n)\ge a^2+b^2+2\sqrt{\dfrac{a^2n}{m}\cdot\dfrac{b^2m}{n}}$
$\to (\dfrac{a^2}{m}+\dfrac{b^2}{n})(m+n)\ge a^2+b^2+2ab$
$\to (\dfrac{a^2}{m}+\dfrac{b^2}{n})(m+n)\ge (a+b)^2$
$\to \dfrac{a^2}{m}+\dfrac{b^2}{n}\ge \dfrac{(a+b)^2}{m+n}$
$\dfrac{a^2}{m} + \dfrac{b^2}{n} \geq \dfrac{(a+b)^2}{m+n}$
$\Leftrightarrow \dfrac{a^2n + b^2m}{mn} \geq \dfrac{a^2 + b^2 + 2ab}{m+ n}$
$\Leftrightarrow (a^2n + b^2m)(m+n) \geq (a^2 +b^2 +2ab)mn$
$\Leftrightarrow a^2mn + a^2n^2 + b^2m^2 + b^2mn \geq a^2mn + b^2mn + 2abmn$
$\Leftrightarrow a^2n^2 – 2abmn + b^2m^2 \geq 0$
$\Leftrightarrow (an – bm)^2 \geq 0$ (luôn đúng)
Vậy $\dfrac{a^2}{m} + \dfrac{b^2}{n} \geq \dfrac{(a+b)^2}{m+n}$