\(\dfrac{\sqrt{ab}}{a+c+b+c}\le\dfrac{\sqrt{ab}}{2\sqrt{\left(a+c\right)\left(b+c\right)}}\le\dfrac{1}{4}\left(\dfrac{a}{a+c}+\dfrac{b}{b+c}\right)\)

Tương tự và cộng lại:

\(A\le\dfrac{1}{4}\left(\dfrac{a}{a+c}+\dfrac{b}{b+c}+\dfrac{b}{a+b}+\dfrac{c}{a+c}+\dfrac{a}{a+b}+\dfrac{c}{b+c}\right)=\dfrac{3}{4}\)

Dấu "=" xảy ra khi \(a=b=c\)

\(\dfrac{ab}{\sqrt{ab+2c}}=\dfrac{ab}{\sqrt{ab+\left(a+b+c\right)c}}=\dfrac{ab}{\sqrt{\left(a+c\right)\left(b+c\right)}}=ab\cdot\sqrt{\dfrac{1}{a+b}\cdot\dfrac{1}{b+c}}\le ab\cdot\dfrac{1}{2}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}\right)=\dfrac{1}{2}\left(\dfrac{ab}{a+b}+\dfrac{ab}{b+c}\right)\)

CMTT: \(\dfrac{bc}{\sqrt{bc+2a}}\le\dfrac{1}{2}\left(\dfrac{bc}{a+b}+\dfrac{bc}{a+c}\right);\dfrac{ac}{\sqrt{ac+2b}}\le\dfrac{1}{2}\left(\dfrac{ac}{b+c}+\dfrac{ac}{b+a}\right)\)

\(\Leftrightarrow P\le\dfrac{1}{2}\left(\dfrac{ab}{c+a}+\dfrac{ab}{c+b}+\dfrac{bc}{b+a}+\dfrac{bc}{c+a}+\dfrac{ac}{b+c}+\dfrac{ac}{b+c}\right)\\ \Leftrightarrow P\le\dfrac{1}{2}\left[\dfrac{b\left(a+c\right)}{a+c}+\dfrac{a\left(b+c\right)}{b+c}+\dfrac{c\left(a+b\right)}{a+b}\right]=\dfrac{1}{2}\left(a+b+c\right)=1\)

Dấu \("="\Leftrightarrow a=b=c=\dfrac{2}{3}\)

\(\dfrac{ab}{\sqrt{ab+2c}}=\dfrac{ab}{\sqrt{ab+c\left(a+b+c\right)}}=\dfrac{ab}{\sqrt{\left(a+c\right)\left(b+c\right)}}\le\dfrac{1}{2}\left(\dfrac{ab}{a+c}+\dfrac{ab}{b+c}\right)\)

Tương tự:

\(\dfrac{bc}{\sqrt{bc+2a}}\le\dfrac{1}{2}\left(\dfrac{bc}{a+b}+\dfrac{bc}{a+c}\right)\) ; \(\dfrac{ca}{\sqrt{ac+2b}}\le\dfrac{1}{2}\left(\dfrac{ca}{a+b}+\dfrac{ca}{b+c}\right)\)

Cộng vế:

\(P\le\dfrac{1}{2}\left(\dfrac{bc}{a+b}+\dfrac{ca}{a+b}+\dfrac{ab}{a+c}+\dfrac{bc}{a+c}+\dfrac{ab}{b+c}+\dfrac{ca}{b+c}\right)=\dfrac{1}{2}\left(a+b+c\right)=1\)

\(P_{max}=1\) khi \(a=b=c=\dfrac{2}{3}\)

Ta có: \(\sqrt{2a+bc}=\sqrt{a^2+ab+ac+bc}=\sqrt{\left(a+b\right)\left(a+c\right)}\le\frac{a+b+a+c}{2}\)

C/m tương tự \(\sqrt{2b+ac}\le\frac{b+a+b+c}{2}\)

                      \(\sqrt{2c+ab}\le\frac{c+a+c+b}{2}\)

\(\Rightarrow Q\le\frac{a+b+a+c+b+a+b+c+c+a+c+b}{2}=\frac{4\left(a+b+c\right)}{2}=4\)

Dấu "=" khi a = b = c = ²/₃

Ớ =( trả lời nhầm nick rồi =(

Chú ý: \(2a^2+ab+2b^2=\frac{5}{4}\left(a+b\right)^2+\frac{3}{4}\left(a-b\right)^2\ge\frac{5}{4}\left(a+b\right)^2\) là ok liền:D

Mấy bạn ơi , cho tớ hỏi:

Luật tính điểm hỏi đáp là gì?
Làm thế nào để câu trả lời của mình đứng đầu tiên trong các câu trả lời?

Ai trả lời nhanh mình tích cho.
 

Câu này giải như sau :

Ta có :

\(\sqrt{2a+bc}=\sqrt{\left(a+b+c\right)a+bc}=\sqrt{a^2+ab+ac+bc}=\sqrt{\left(a+c\right)\left(a+b\right)}\)

\(\Rightarrow\sqrt{2a+bc}\le\frac{a+b+a+c}{2}=\frac{2a+b+c}{2}\left(1\right)\)

tương tự ta có :\(\sqrt{2b+ac}\le\frac{2b+a+c}{2}\left(2\right)\)

\(\sqrt{2c+ac}\le\frac{2c+a+c}{2}\left(3\right)\)

cộng vế với vế 1,2,3 ta được

\(Q\le\frac{3\left(a+b+c\right)}{2}=\frac{3.2}{2}=3\)\(\Rightarrow Q_{max}=3\Leftrightarrow\)dấu "=" (a,b,c) là hoán vị của \(\left(0.1.1\right)\)

@Hoàng Thanh Tuấn bạn giải sai rồi 

\(\sqrt{2a^2+ab+2b^2}=\sqrt{\dfrac{3}{2}\left(a^2+b^2\right)+\dfrac{1}{2}\left(a+b\right)^2}\ge\sqrt{\dfrac{3}{4}\left(a+b\right)^2+\dfrac{1}{2}\left(a+b\right)^2}=\dfrac{\sqrt{5}}{2}\left(a+b\right)\)

Tương tự:

\(\sqrt{2b^2+bc+2c^2}\ge\dfrac{\sqrt{5}}{2}\left(b+c\right)\) ; \(\sqrt{2c^2+ca+2a^2}\ge\dfrac{\sqrt{5}}{2}\left(c+a\right)\)

Cộng vế với vế:

\(P\ge\sqrt{5}\left(a+b+c\right)\ge\dfrac{\sqrt{5}}{3}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^3=\dfrac{\sqrt{5}}{3}\)

Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{9}\)