Ta có:

\(\left(2a^2-b^2-c^2\right)^2\ge0\)

\(\Leftrightarrow4a^4+b^4+c^4-4a^2b^2-4a^2c^2+2b^2c^2\ge0\)

\(\Leftrightarrow a^4+b^4+c^4+2a^2b^2+2b^2c^2+2c^2a^2\ge6a^2b^2+6a^2c^2-3a^4\)

\(\Leftrightarrow\left(a^2+b^2+c^2\right)^2\ge3a^2\left(2b^2+2c^2-a^2\right)\)

\(\Leftrightarrow\dfrac{1}{\sqrt{2b^2+2c^2-a^2}}\ge\dfrac{\sqrt{3}a}{a^2+b^2+c^2}\)

\(\Leftrightarrow\dfrac{a}{\sqrt{2b^2+2c^2-a^2}}\ge\sqrt{3}\dfrac{a^2}{a^2+b^2+c^2}\)

Tương tự: \(\dfrac{b}{\sqrt{2a^2+2c^2-b^2}}\ge\sqrt{3}.\dfrac{b^2}{a^2+b^2+c^2}\) ; \(\dfrac{c}{\sqrt{2a^2+2b^2-c^2}}\ge\sqrt{3}.\dfrac{c^2}{a^2+b^2+c^2}\)

Cộng vế: \(P\ge\dfrac{\sqrt{3}\left(a^2+b^2+c^2\right)}{a^2+b^2+c^2}=\sqrt{3}\)

\(P_{min}=\sqrt{3}\) khi \(a=b=c\)

Tham khảo ạ !

t nhớ Akai Haruma làm bài này rồi.CHTT đi:v

Áp dụng bđt Cauchy Shwarz và bđt phụ \(\dfrac{1}{x+y}\le\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)

\(\Rightarrow M^2=\left(\sqrt{\dfrac{a}{b+c+2a}}+\sqrt{\dfrac{b}{c+a+2b}}+\sqrt{\dfrac{c}{a+b+2c}}\right)^2\)

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

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

\(=\dfrac{3}{4}\left(\dfrac{a+b}{a+b}+\dfrac{b+c}{b+c}+\dfrac{c+a}{c+a}\right)=\dfrac{9}{4}\)

➤ \(M\le\dfrac{3}{2}\)

Dấu "=" xảy ra ⇔ a = b = c

\(M=\sqrt{\dfrac{a}{b+c+2a}}+\sqrt{\dfrac{b}{c+a+2b}}+\sqrt{\dfrac{c}{a+b+2c}}\)

\(\le\dfrac{1}{4}+\dfrac{a}{b+c+2a}+\dfrac{1}{4}+\dfrac{b}{c+a+2b}+\dfrac{1}{4}+\dfrac{c}{a+b+2c}\)

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

\(=\dfrac{3}{4}+\dfrac{1}{4}.\left(1+1+1\right)=\dfrac{3}{2}\)

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

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 \(a=b=c=\dfrac{1}{9}\)

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

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 \(a=b=c\)