Áp dụng BĐT AM-GM ta có:

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

Tương tự ta cũng có:\(\sqrt{b\left(c+2a\right)}\le\frac{3b+c+2a}{2\sqrt{3}}\)

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

Cộng theo vế các BĐT lại ta được:

\(VT\le\frac{3a+b+2c}{2\sqrt{3}}+\frac{3b+c+2a}{2\sqrt{3}}+\frac{3c+a+2b}{2\sqrt{3}}=\frac{6a+6b+6c}{2\sqrt{3}}=\frac{6.4}{2\sqrt{3}}=4\sqrt{3}\)

Dấu "=" xảy ra khi \(a=b=c=\frac{4}{3}\)

có cả mấy bất đẳng thức đó hả

bn viết công thức tổng quát ra cho mk vs

mk thanks

Nè bạn :) 

Ta có : \(2ab+2ac\ge4a\sqrt{bc}\) (Cauchy_)

\(\Rightarrow a^2+2ab+2ac+4bc\ge a^2+4a\sqrt{bc}+4bc\)

\(\Rightarrow a^2+2ab+2ac+4bc\ge\left(a+2\sqrt{bc}\right)^2\)

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

Tương tự : \(\sqrt{\left(b+2a\right)\left(b+2c\right)}\ge b+2\sqrt{ac}\)\(\left(2\right)\)

\(\sqrt{\left(c+2a\right)\left(c+2b\right)}\ge c+2\sqrt{ab}\)\(\left(3\right)\)

Từ \(\left(1\right);\left(2\right);\left(3\right)\)\(\Rightarrow\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2\ge3\)

\(\Rightarrow\sqrt{a}+\sqrt{b}+\sqrt{c}\ge\sqrt{3}\)

Đẳng thức xảy ra khi \(a=b=c=\frac{1}{\sqrt{3}}\)

Thay vào biểu thức M ta được M = \(\frac{\sqrt{3}}{3}\)

tth làm nhanh a đang cần =)))

1)

Ta có: \(M=\Sigma_{cyc}\frac{\sqrt{3}\left(a+b+4c\right)}{\sqrt{3\left(a+b\right)\left(a+b+4c\right)}}\ge\Sigma_{cyc}\frac{\sqrt{3}\left(a+b+4c\right)}{\frac{3\left(a+b\right)+\left(a+b+4c\right)}{2}}=\Sigma_{cyc}\frac{\sqrt{3}\left(a+b+4c\right)}{2\left(a+b+c\right)}=3\sqrt{3}\)

Dấu "=" xảy ra khi a=b=c

2)

\(\Sigma_{cyc}\sqrt[3]{\left(\frac{2a}{ab+1}\right)^2}=\Sigma_{cyc}\frac{2a}{\sqrt[3]{2a\left(ab+1\right)^2}}\ge\Sigma_{cyc}\frac{2a}{\frac{2a+\left(ab+1\right)+\left(ab+1\right)}{3}}=3\Sigma_{cyc}\frac{a}{ab+a+1}\)

Ta có bổ đề: \(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1}=1\left(abc=1\right)\)

\(\Rightarrow\Sigma_{cyc}\sqrt[3]{\left(\frac{2a}{ab+1}\right)^2}\ge3\)

b+c\(\ge\) \(2\sqrt{bc}\)

(a+2b)(a+2c) =\(a^2 +2ac+2ab+ 4bc= a^2+2a(b+c) +4bc\)

\(\ge\)\(a^2+4a.\sqrt{bc}+4bc=\left(a+2\sqrt{bc}\right)^2\)

\(=>\sqrt{\left(a+2b\right)\left(a+2c\right)}=a+2\sqrt{bc}\)

tương tự: \(\sqrt{\left(b+2a\right)\left(b+2c\right)}=b+2\sqrt{ac}\)

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

\(=>\sqrt{\left(a+2b\right)\left(a+2c\right)}+\sqrt{\left(b+2a\right)\left(b+2c\right)}+\sqrt{\left(c+2b\right)\left(c+2a\right)}\ge a+b+c+2\sqrt{ab}+2\sqrt{bc}+2\sqrt{ac}=\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2=3\)

khi a=b=c ( a,b,c nguyên dương nên a+b+c>0)

=> \(3\sqrt{a}=\sqrt{3}=>\sqrt{a}=\sqrt{b}=\sqrt{c}=\dfrac{\sqrt{3}}{3}\)

Thay vào M=\(\dfrac{1}{3}\)