ý sai rồi . vậy mới đúng

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

ko cần nữa

Áp dụng BĐT Cauchy-Schwarz ta có:

\(VT=\dfrac{1}{\sqrt{a}}+\dfrac{3}{\sqrt{b}}+\dfrac{8}{\sqrt{3c+2a}}\)

\(=\dfrac{1}{\sqrt{a}}+\dfrac{1}{\sqrt{b}}+\dfrac{2}{\sqrt{b}}+\dfrac{8}{\sqrt{3c+2a}}\)

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

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

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

\(\ge\dfrac{64}{\sqrt{\left(1+2^2+3\right)\left(a+2a+3c+3b\right)}}\)

\(=\dfrac{64}{\sqrt{24\left(a+c+b\right)}}=\dfrac{16\sqrt{2}}{\sqrt{3\left(a+b+c\right)}}=VP\)

\(=\dfrac{\sqrt{ab}}{b}+\sqrt{\dfrac{a^2b}{b^2a}}=\dfrac{\sqrt{ab}}{b}+\sqrt{\dfrac{a}{b}}=\dfrac{\sqrt{ab}}{b}+\dfrac{\sqrt{ab}}{b}=\dfrac{2\sqrt{ab}}{b}\left(B\right)\)

B

B

Đặt \(\left(\sqrt{a};\sqrt{b};\sqrt{c}\right)=\left(x;y;z\right)\Rightarrow x+y+z=1\)

BĐT trở thành: \(\dfrac{xy}{\sqrt{x^2+y^2+2z^2}}+\dfrac{yz}{\sqrt{y^2+z^2+2x^2}}+\dfrac{zx}{\sqrt{x^2+z^2+2y^2}}\le\dfrac{1}{2}\)

Ta có:

\(x^2+z^2+y^2+z^2\ge\dfrac{1}{2}\left(x+z\right)^2+\dfrac{1}{2}\left(y+z\right)^2\ge\left(x+z\right)\left(y+z\right)\)

\(\Rightarrow\dfrac{xy}{\sqrt{x^2+y^2+2z^2}}\le\dfrac{xy}{\sqrt{\left(x+z\right)\left(y+z\right)}}\le\dfrac{1}{2}\left(\dfrac{xy}{x+z}+\dfrac{xy}{y+z}\right)\)

Tương tự: \(\dfrac{yz}{\sqrt{y^2+z^2+2x^2}}\le\dfrac{1}{2}\left(\dfrac{yz}{x+y}+\dfrac{yz}{x+z}\right)\)

\(\dfrac{zx}{\sqrt{z^2+x^2+2y^2}}\le\dfrac{1}{2}\left(\dfrac{zx}{x+y}+\dfrac{zx}{y+z}\right)\)

Cộng vế với vế:

\(VT\le\dfrac{1}{2}\left(\dfrac{zx+yz}{x+y}+\dfrac{xy+zx}{y+z}+\dfrac{yz+xy}{z+x}\right)=\dfrac{1}{2}\left(x+y+z\right)=\dfrac{1}{2}\) (đpcm)

Dấu "=" xảy ra khi \(x=y=z\) hay \(a=b=c\)

Áp dụng BĐT: \(x^2+y^2+z^2\ge\dfrac{1}{3}\left(x+y+z\right)^2\) ta có:

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

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

Cộng vế với vế và rút gọn:

\(\sqrt{\dfrac{a+2b}{3}}+\sqrt{\dfrac{b+2c}{3}}+\sqrt{\dfrac{c+2a}{3}}\ge\sqrt{a}+\sqrt{b}+\sqrt{c}\) (đpcm)

\(ab\cdot\sqrt{\dfrac{a}{3b}}-a^2\sqrt{\dfrac{3b}{a}}\)

\(=a\sqrt{ab}-a^2\cdot\dfrac{\sqrt{3b}}{\sqrt{a}}\)

\(=a\sqrt{ab}-a\sqrt{a}\cdot\sqrt{3b}\)

\(=a\sqrt{ab}\left(1-\sqrt{3}\right)\)

\(\Leftrightarrow m=\dfrac{a\sqrt{ab}\left(1-\sqrt{3}\right)}{\sqrt{3ab}}=\dfrac{a\left(\sqrt{3}-3\right)}{3}\)