Question

[Hello-Tôi yêu các bạn]

a,b,c>0, biết a+b+c=3

CMR a)\(\frac{ab}{\sqrt{a^2+3b^2}}+\frac{bc}{\sqrt{b^2+3c^2}}+\frac{ac}{\sqrt{c^2+3a^2}}\)≤\(\frac{3}{2}\)

b)\(\frac{a}{\sqrt{b^2+3}}+\frac{b}{\sqrt{c^2+3}}+\frac{c}{\sqrt{a^2+3}}\)≥\(\frac{3}{2}\)

Answer 1

[Phạm Minh Quang]

b) Ta có:

\(\frac{a}{\sqrt{b^2+3}}+\frac{a}{\sqrt{b^2+3}}+\frac{b^2+3}{8}+\frac{a^2}{2}\)\(\ge\)\(4\sqrt[4]{\frac{a^4}{16}}=2a\)

\(\frac{b}{\sqrt{c^2+3}}+\frac{b}{\sqrt{c^2+3}}+\frac{c^2+3}{8}+\frac{b^2}{2}\ge4\sqrt[4]{\frac{b^4}{16}}=2b\)

\(\frac{c}{\sqrt{a^2+3}}+\frac{c}{\sqrt{a^2+3}}+\frac{a^2+3}{8}+\frac{c^2}{2}\ge4\sqrt[4]{\frac{c^4}{16}}=2c\)

Cộng lại ta đươc:

\(2\left(\frac{a}{\sqrt{b^2+3}}+\frac{b}{\sqrt{c^2+3}}+\frac{c}{\sqrt{a^2+3}}\right)+\)\(\frac{5\left(a^2+b^2+c^2\right)+9}{8}\)\(\ge2\left(a+b+c\right)\)

⇒ \(2\left(\frac{a}{\sqrt{b^2+3}}+\frac{b}{\sqrt{c^2+3}}+\frac{c}{\sqrt{a^2+3}}\right)\ge\)\(6-\frac{5\left(a^2+b^2+c^2\right)+9}{8}\)(1)

Lại có: \(a^2+1\ge2a\); \(b^2+1\ge2b\); \(c^2+1\ge2c\)

Suy ra \(a^2+b^2+c^2\ge2\left(a+b+c\right)-3=3\)

Khi đó (1)⇔ \(2\left(\frac{a}{\sqrt{b^2+3}}+\frac{b}{\sqrt{c^2+3}}+\frac{c}{\sqrt{a^2+3}}\right)\ge\)\(6-\frac{5.3+9}{8}=3\)

⇒ \(\frac{a}{\sqrt{b^2+3}}+\frac{b}{\sqrt{c^2+3}}+\frac{c}{\sqrt{a^2+3}}\ge\frac{3}{2}\)

Dấu "=" xảy ra ⇔ \(a=b=c=1\)

Answer 2

[Nguyễn Việt Lâm]

\(\left(a^2+3b^2\right)\left(1+3\right)\ge\left(a+3b\right)^2\Rightarrow\sqrt{a^2+3b^2}\ge\frac{a+3b}{2}\)

\(\Rightarrow P=\sum\frac{ab}{\sqrt{a^2+3b^2}}\le2\sum\frac{ab}{a+3b}=2\sum\frac{ab}{a+b+b+b}\)

\(\Rightarrow P\le\frac{1}{8}\sum ab\left(\frac{1}{a}+\frac{3}{b}\right)=\frac{1}{8}\sum\left(3a+b\right)=\frac{1}{2}\left(a+b+c\right)=\frac{3}{2}\)

"=" \(\Leftrightarrow a=b=c=1\)