có cả mấy bất đẳng thức đó hả

bn viết công thức tổng quát ra cho mk vs

mk thanks

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=\(\dfrac{1}{3}\)

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 = \(\frac{\sqrt{3}}{3}\)

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 \(a=b=c=\frac{4}{3}\)

Câu b tương tự

a/ \(\sqrt[5]{2a+b}+\sqrt[5]{2b+c}+\sqrt[5]{2c+a}\)

\(=\frac{1}{\sqrt[5]{3^4}}\left(\sqrt[5]{3^4}.\sqrt[5]{2a+b}+\sqrt[5]{3^4}.\sqrt[5]{2b+c}+\sqrt[5]{3^4}.\sqrt[5]{2c+a}\right)\)

\(\le\frac{1}{\sqrt[5]{3^4}}\left(\frac{3+3+3+3+2a+b}{5}+\frac{3+3+3+3+2b+c}{5}+\frac{3+3+3+3+2c+a}{5}\right)\)

\(=\frac{1}{\sqrt[5]{3^4}}\left(\frac{36}{5}+\frac{3\left(a+b+c\right)}{5}\right)\)

\(=\frac{1}{\sqrt[5]{3^4}}.9=3\sqrt[5]{3}\)

b/ \(\sqrt[5]{a\left(a+c\right)\left(2a+b\right)}+\sqrt[5]{b\left(b+a\right)\left(2b+c\right)}+\sqrt[5]{c\left(c+b\right)\left(2c+a\right)}\)

\(\frac{1}{\sqrt[5]{6^4}}.\left(\sqrt[5]{6^2}.\sqrt[5]{6.a.3.\left(a+c\right).2.\left(2a+b\right)}+\sqrt[5]{6^2}.\sqrt[5]{6.b.3.\left(b+a\right).2.\left(2b+c\right)}+\sqrt[5]{6^2}.\sqrt[5]{6.c.3.\left(c+b\right).2.\left(2c+a\right)}\right)\)

\(\le\frac{1}{\sqrt[5]{6^4}}.\left(\frac{6+6+6a+3\left(a+c\right)+2\left(2a+b\right)}{5}+\frac{6+6+6b+3\left(b+a\right)+2\left(2b+c\right)}{5}+\frac{6+6+6c+3\left(c+b\right)+2\left(2c+a\right)}{5}\right)\)

\(=\frac{1}{\sqrt[5]{6^4}}.\left(\frac{36}{5}+\frac{18\left(a+b+c\right)}{5}\right)\)

\(=\frac{1}{\sqrt[5]{6^4}}.18=3\sqrt[5]{6}\)

\(\sqrt{a^2+2ac+2ab+4bc}\) + \(\sqrt{b^2+2bc+2ab+4ac}\) + \(\sqrt{c^2+2bc+2ac+4ab}\) =3

Haizzz mọi người ra chưa?