Question

Chứng minh:

a) \(\dfrac{a^2+3}{\sqrt{a^2+2}}>2\)

b) \(\sqrt{a}+\sqrt{b}\le\dfrac{a}{\sqrt{b}}+\dfrac{b}{\sqrt{a}}\) với a > 0; b > 0

Answer 1

a: \(\dfrac{a^2+3}{\sqrt{a^2+2}}=\dfrac{a^2+2+1}{\sqrt{a^2+2}}=\sqrt{a^2+2}+\dfrac{1}{\sqrt{a^2+2}}>2\cdot\sqrt{\sqrt{a^2+2}\cdot\dfrac{1}{\sqrt{a^2+2}}}=2\)

b: \(\Leftrightarrow\left(\sqrt{a}+\sqrt{b}\right)\cdot\sqrt{ab}< =a\sqrt{a}+b\sqrt{b}\)

\(\Leftrightarrow a\sqrt{b}+b\sqrt{a}-a\sqrt{a}-b\sqrt{b}< =0\)

\(\Leftrightarrow a\left(\sqrt{b}-\sqrt{a}\right)-b\left(\sqrt{b}-\sqrt{a}\right)< =0\)

\(\Leftrightarrow\left(a-b\right)\left(\sqrt{b}-\sqrt{a}\right)< =0\)(luôn đúng)