Question

CMR với mọi a,b > 0:

\(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\).

Answer 1

\(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\Rightarrow ab+b^2+a^2+ab\ge4ab\left(a,b>0\right)\)

<=>a²+b²-2ab\(\ge\)0

<=>(a-b)²\(\ge\)0(luôn đúng)

=>điều cần chứng minh

Answer 2

\(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)

\(\frac{a+b}{ab}\ge\frac{4}{a+b}\)

(a + b) (a + b) \(\ge\) 4ab

\(\Rightarrow\left(a+b\right)^2\ge4ab\)

Mà a,b > 0 nên a + b > 0 

=> \(\left(a+b\right)^2\ge4ab\)