Question

Cho a , b >0 thỏa mãn \(\dfrac{1}{a^2}+\dfrac{1}{b^2}=2\)

Chứng minh : a +b ≥ 2

Answer 1

Lời giải:

Áp dụng BĐT Bunhiacopxky:

\(\left(\frac{1}{a^2}+\frac{1}{b^2}\right)(1+1)\geq \left(\frac{1}{a}+\frac{1}{b}\right)^2\)

\(\left(\frac{1}{a}+\frac{1}{b}\right)(a+b)\geq (1+1)^2\Rightarrow \frac{1}{a}+\frac{1}{b}\geq \frac{4}{a+b}\)

Do đó:

\(2(\frac{1}{a^2}+\frac{1}{b^2})\geq (\frac{4}{a+b})^2\)

\(\Leftrightarrow 4\geq (\frac{4}{a+b})^2\)

\(\Rightarrow 2\geq \frac{4}{a+b}(\forall a,b>0)\Rightarrow 2(a+b)\geq 4\Rightarrow a+b\geq 2\) (đpcm)

Dấu bằng xảy ra khi $a=b=1$