Áp dụng BĐt bunhiakovsky ta có:

`(\sqrt{a(3a+b)}+\sqrt{b(3b+a)})^2<=(a+b)(3a+b+3b+a)`

`<=>(\sqrt{a(3a+b)}+\sqrt{b(3b+a)})^2<=4(a+b)^2`

`<=>\sqrt{a(3a+b)}+\sqrt{b(3b+a)}<=2(a+b)`

`=>(a+b)/(\sqrt{a(3a+b)}+\sqrt{b(3b+a)})>=1/2`

Dấu "=" `<=>a=b`

Ta có \(\sqrt{2022a+\dfrac{\left(b-c\right)^2}{2}}\) 

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

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

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

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

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

Vậy \(\sqrt{2022a+\dfrac{\left(b-c\right)^2}{2}}\le\dfrac{2a+b+c}{\sqrt{2}}\). Lập 2 BĐT tương tự rồi cộng vế, ta được \(VT\le\dfrac{2a+b+c+2b+c+a+2c+a+b}{\sqrt{2}}\)

\(=\dfrac{4\left(a+b+c\right)}{\sqrt{2}}\) \(=\dfrac{4.1011}{\sqrt{2}}\) \(=2022\sqrt{2}\)

ĐTXR \(\Leftrightarrow\) \(\left\{{}\begin{matrix}ab=0\\bc=0\\ca=0\\a+b+c=1011\end{matrix}\right.\) \(\Leftrightarrow\left(a;b;c\right)=\left(1011;0;0\right)\) hoặc các hoán vị. Vậy ta có đpcm.

Dễ dàng c/m : \(\dfrac{1}{a+2}+\dfrac{1}{b+2}+\dfrac{1}{c+2}=1\)

Ta có : \(\dfrac{1}{\sqrt{2\left(a^2+b^2\right)}+4}\le\dfrac{1}{a+b+4}\le\dfrac{1}{4}\left(\dfrac{1}{a+2}+\dfrac{1}{b+2}\right)\) 

Suy ra : \(\Sigma\dfrac{1}{\sqrt{2\left(a^2+b^2\right)}+4}\le2.\dfrac{1}{4}\left(\dfrac{1}{a+2}+\dfrac{1}{b+2}+\dfrac{1}{c+2}\right)=\dfrac{1}{2}.1=\dfrac{1}{2}\) 

" = " \(\Leftrightarrow a=b=c=1\)

Ta có:

\(\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2=9\\ \Leftrightarrow a+b+c+2\sqrt{ab}+2\sqrt{bc}+2\sqrt{ac}=9\\ \Leftrightarrow\sqrt{ab}+\sqrt{bc}+\sqrt{ac}=2\)

\(\Rightarrow\dfrac{\sqrt{a}}{a+2}+\dfrac{\sqrt{b}}{b+2}+\dfrac{\sqrt{c}}{c+2}=\dfrac{\sqrt{a}}{a+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}}+\dfrac{\sqrt{b}}{b+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}}+\dfrac{\sqrt{c}}{c+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}}\\ =\dfrac{\sqrt{a}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)}+\dfrac{\sqrt{b}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{b}+\sqrt{c}\right)}+\dfrac{\sqrt{c}}{\left(\sqrt{b}+\sqrt{c}\right)\left(\sqrt{a}+\sqrt{c}\right)}\\ =\dfrac{\sqrt{a}\left(\sqrt{b}+\sqrt{c}\right)+\sqrt{b}\left(\sqrt{a}+\sqrt{c}\right)+\sqrt{c}\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{b}+\sqrt{c}\right)\left(\sqrt{a}+\sqrt{c}\right)}\\ =\dfrac{2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\right)}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{b}+\sqrt{c}\right)\left(\sqrt{a}+\sqrt{c}\right)}\\ =\dfrac{4}{\sqrt{\left(\sqrt{a}+\sqrt{b}\right)^2\left(\sqrt{b}+\sqrt{c}\right)^2\left(\sqrt{a}+\sqrt{c}\right)^2}}\)\(=\dfrac{4}{\sqrt{\left(a+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\right)\left(b+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\right)\left(c+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\right)}}\\ =\dfrac{4}{\sqrt{\left(a+2\right)\left(b+2\right)\left(c+2\right)}}\)

hello nha

2k? vậy ạ

Áp dụng BĐT Bunyakovsky dạng cộng mẫu:

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

\(=\frac{\left(-\sqrt{c}\right)^2}{\sqrt{b}+\sqrt{c}}=\frac{c}{\sqrt{b}+\sqrt{c}}\)

Tương tự CM được: \(4\left[\frac{\left(\sqrt{a}-1\right)^2}{\sqrt{b}}+\frac{\left(\sqrt{b}-1\right)^2}{\sqrt{c}}+\frac{\left(\sqrt{c}-1\right)^2}{\sqrt{a}}\right]\ge2\left(\frac{a}{\sqrt{c}+\sqrt{a}}+\frac{b}{\sqrt{a}+\sqrt{b}}+\frac{c}{\sqrt{b}+\sqrt{c}}\right)\) (1)

Lại có: \(VP\left(1\right)-\left(\frac{a+b}{\sqrt{a}+\sqrt{b}}+\frac{b+c}{\sqrt{b}+\sqrt{c}}+\frac{c+a}{\sqrt{c}+\sqrt{a}}\right)=...=0\) (biến đổi đồng nhất)

=> \(VT\left(1\right)\ge\frac{a+b}{\sqrt{a}+\sqrt{b}}+\frac{b+c}{\sqrt{b}+\sqrt{c}}+\frac{c+a}{\sqrt{c}+\sqrt{a}}\)

Dấu "=" xảy ra khi: \(a=b=c=\frac{4}{9}\)

Ai giải được không ?