Question

Cho a,b,c > 1 và \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=2\).CMR:

\(\sqrt{a-1}+\sqrt{b-1}+\sqrt{c-1}\le\sqrt{a+b+c}\)

Answer 1

Lời giải:

Áp dụng BĐT Bunhiacopxky ta có:

\((\sqrt{a-1}+\sqrt{b-1}+\sqrt{c-1})^2\leq \left(\frac{a-1}{a}+\frac{b-1}{b}+\frac{c-1}{c}\right)(a+b+c)\)

\(\Leftrightarrow (\sqrt{a-1}+\sqrt{b-1}+\sqrt{c-1})^2\leq \left(3-\frac{1}{a}-\frac{1}{b}-\frac{1}{c}\right)(a+b+c)\)

\(\Leftrightarrow (\sqrt{a-1}+\sqrt{b-1}+\sqrt{c-1})^2\leq \left(3-2\right)(a+b+c)\)

\(\Leftrightarrow (\sqrt{a-1}+\sqrt{b-1}+\sqrt{c-1})^2\leq a+b+c\)

\(\Leftrightarrow \sqrt{a-1}+\sqrt{b-1}+\sqrt{c-1}\leq \sqrt{a+b+c}\)

Ta có đpcm.

Dấu bằng xảy ra khi \(a=b=c=\frac{3}{2}\)