Question

Cho hai số dương a,b thỏa mãn \(\frac{1}{a^2}+\frac{1}{b^2}=2\) . Cmr \(a+b\ge2\)

Answer 1

Lời giải:

Áp dụng BĐT Bunhiacopxky cho $a,b$ dương:

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

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

Do đó:

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

\(\Rightarrow 4(a+b)^2\geq 16\Rightarrow (a+b)^2\geq 4\Rightarrow a+b\geq 2\) (đpcm)

Dấu "=" xảy ra khi $a=b=1$