bđt phụ sai mà cũng ko đc chuẩn hóa

\(\frac{ab}{a^2+b^2}\le\frac{ab}{2ab}=\frac{1}{2}\)

tương tự \(\frac{\Rightarrow ab}{a^2+b^2}+\frac{bc}{b^2+c^2}+\frac{ac}{a^2+c^2}\le\frac{3}{2}\)

=>Thắng Nguyễn :cm theo cách đó sai

SOS cho khỏe :v 

WLOG \(a\ge b\ge c\)

Áp dụng BĐT AM-GM ta có:

\(b^2Σ_{cyc}\left(a^3+\frac{4ab}{a^2+b^2}-3\right)=b^2\left(Σ_{cyc}(a^3-abc)-2Σ_{cyc}\left(1-\frac{2ab}{a^2+b^2}\right)\right)\)

\(=b^2Σ_{cyc}(a-b)^2\left(\frac{a+b+c}{2}-\frac{2}{a^2+b^2}\right)=\frac{b^2}{2}Σ_{cyc}\frac{(a-b)^2((a+b+c)(a^2+b^2)-4abc)}{a^2+b^2}\)

\(\ge\frac{b^2}{2}Σ_{cyc}\frac{(a-b)^2((a+b+c)2ab-4abc)}{a^2+b^2}=b^2Σ_{cyc}\frac{(a-b)^2ab(a+b-c)}{a^2+b^2}\)

\(\ge\frac{b^2(a-c)^2ac(a+c-b)}{a^2+c^2}+\frac{b^2(b-c)^2bc(b+c-a)}{b^2+c^2}\)

\(\ge\frac{a^2(b-c)^2ac(a-b)}{a^2+c^2}+\frac{b^2(b-c)^2bc(b-a)}{b^2+c^2}\)

\(=\frac{abc^3(a+b)(b-c)^2(a-b)^2}{(a^2+c^2)(b^2+c^2)}\ge0\) (đúng :v)

Ta có 

\(\sqrt{\frac{ab}{c+ab}}=\sqrt{\frac{ab}{c\left(a+b+c\right)+ab}}\)\(=\sqrt{\frac{ab}{\left(c+a\right)\left(c+b\right)}}\)\(=\sqrt{\frac{a}{c+a}}.\sqrt{\frac{b}{c+b}}\)\(\le\frac{1}{2}\left(\frac{a}{c+a}+\frac{b}{c+b}\right)\)

Tương tự, ta có

\(\sqrt{\frac{bc}{a+bc}}\le\frac{1}{2}\left(\frac{b}{a+b}+\frac{c}{a+c}\right)\)

\(\sqrt{\frac{ca}{b+ca}\le\frac{1}{2}\left(\frac{c}{c+b}+\frac{a}{b+a}\right)}\)

Cộng vế theo vế của 3 bđt ta được đpcm

BĐT <=> \(\frac{2}{a^2+2}+\frac{2}{b^2+2}+\frac{2}{c^2+2}\le2\)

\(\Leftrightarrow1-\frac{a^2}{a^2+2}+1-\frac{b^2}{b^2+2}+1-\frac{c^2}{c^2+2}\le2\)

\(\Leftrightarrow\frac{a^2}{a^2+2}+\frac{b^2}{b^2+2}+\frac{c^2}{c^2+2}\ge1\)

Theo BĐT Svacxo:

\(VT\ge\frac{\left(a+b+c\right)^2}{a^2+b^2+c^2+6}=\frac{a^2+b^2+c^2+2\left(ab+bc+ca\right)}{a^2+b^2+c^2+6}=\frac{a^2+b^2+c^2+6}{a^2+b^2+c^2+6}=1\)

Vậy ta có đpcm.

P/s: Đúng ko ta?

Mk cx đang định hỏi câu này

Ta có \(ab+bc+ca\ge3\sqrt[3]{a^2b^2c^2}\)\(\Rightarrow3\sqrt[3]{a^2b^2c^2}\le3\Leftrightarrow abc\le1\)

\(\Rightarrow\)\(\frac{1}{1+a^2\left(b+c\right)}\le\frac{1}{abc+a^2\left(b+c\right)}\)\(=\frac{1}{a\left(ab+bc+ca\right)}=\frac{1}{3a}\)

\(CMTT\Rightarrow\frac{1}{1+b^2\left(c+a\right)}\le\frac{1}{3b}\)

                  \(\frac{1}{1+c^2\left(a+b\right)}\le\frac{1}{3c}\)

\(\Rightarrow VT\le\frac{1}{3a}+\frac{1}{3b}+\frac{1}{3c}\)\(=\frac{ab+bc+ca}{3abc}=\frac{1}{abc}\)