Question

Cho a, b, c, d là những số dương. 

Chứng minh rằng :             

                 \(\dfrac{a}{\sqrt{b}}+\dfrac{b}{\sqrt{a}}\ge\sqrt{a}+\sqrt{b}\)

Answer 1

Ta có: \(\dfrac{a}{\sqrt{b}}+\dfrac{b}{\sqrt{a}}\)\(\Leftrightarrow\dfrac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}\sqrt{b}}\).
Giả sử: \(\dfrac{a}{\sqrt{b}}+\dfrac{b}{\sqrt{a}}\ge\sqrt{a}+\sqrt{b}\)\(\Leftrightarrow\dfrac{a\sqrt{a}+b\sqrt{b}}{\sqrt{ab}}\ge\sqrt{a}+\sqrt{b}\)
\(\Leftrightarrow a\sqrt{a}+b\sqrt{b}=\left(\sqrt{a}+\sqrt{b}\right)\sqrt{ab}\)
\(\Leftrightarrow\left(\sqrt{a}+\sqrt{b}\right)\left(a+b-\sqrt{ab}\right)\ge\left(\sqrt{a}+\sqrt{b}\right)\sqrt{ab}\)
\(\Leftrightarrow a+b-\sqrt{ab}\ge\sqrt{ab}\)\(\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2\ge0\) (Luôn đúng).
Vì vậy: \(\dfrac{a}{\sqrt{b}}+\dfrac{b}{\sqrt{a}}\ge\sqrt{a}+\sqrt{b}\).

Answer 2

giải

áp dụng bđt cauchy-schwarz ta có

\(\left\{{}\begin{matrix}\dfrac{a}{\sqrt{b}}+\sqrt{b}\ge2\sqrt{a}\\\dfrac{b}{\sqrt{a}}+\sqrt{a}\ge2\sqrt{b}\end{matrix}\right.\)

\(\Rightarrow\dfrac{a}{\sqrt{b}}+\sqrt{b}+\dfrac{b}{\sqrt{a}}+\sqrt{a}\ge2\sqrt{a}+2\sqrt{b}\)

\(\Leftrightarrow\dfrac{a}{\sqrt{b}}+\dfrac{b}{\sqrt{a}}\ge\sqrt{a}+\sqrt{b}\)

\(\Rightarrow dpcm\)