Question

[Lưu Thị Thảo Ly]

cmr\(\dfrac{a+b}{\sqrt{a\left(3a+b\right)}+\sqrt{b\left(3b+a\right)}}\ge\dfrac{1}{2}\)

(a,b>0)

Answer 1

[Akai Haruma]

Lời giải:

Áp dụng BĐT Cauchy:

\(2\sqrt{a(3a+b)}=\sqrt{4a(3a+b)}\leq \frac{4a+3a+b}{2}\)

Tương tự \(2\sqrt{b(3b+a)}\leq \frac{4b+3b+a}{2}\)

\(\Rightarrow 2(\sqrt{a(3a+b)}+\sqrt{b(3b+a)})\leq \frac{8a+8b}{2}=4(a+b)\)

\(\Rightarrow \sqrt{a(3a+b)}+\sqrt{b(3b+a)}\leq 2(a+b)\)

\(\Rightarrow \frac{a+b}{\sqrt{a(3a+b)}+\sqrt{b(3b+a)}}\geq \frac{a+b}{2(a+b)}=\frac{1}{2}\) (đpcm)

Dấu bằng xảy ra khi \(a=b>0\)