Trước hết theo BĐT Schur bậc 3 ta có:

\(\left(a+b+c\right)\left(a^2+b^2+c^2\right)+9abc\ge2\left(a+b+c\right)\left(ab+bc+ca\right)\)

\(\Leftrightarrow a^2+b^2+c^2+3abc\ge2\left(ab+bc+ca\right)\) (do \(a+b+c=3\)) (1)

Đặt vế trái BĐT cần chứng minh là P, ta có:

\(P=\dfrac{\left(a^2+abc\right)^2}{a^2b^2+2abc^2}+\dfrac{\left(b^2+abc\right)^2}{b^2c^2+2a^2bc}+\dfrac{\left(c^2+abc\right)^2}{a^2c^2+2ab^2c}\)

\(\Rightarrow P\ge\dfrac{\left(a^2+b^2+c^2+3abc\right)^2}{a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)}=\dfrac{\left(a^2+b^2+c^2+3abc\right)^2}{\left(ab+bc+ca\right)^2}\)

Áp dụng (1):

\(\Rightarrow P\ge\dfrac{\left[2\left(ab+bc+ca\right)\right]^2}{\left(ab+bc+ca\right)^2}=4\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c=1\)

Từ \(7\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)=6\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)+2017\)

\(\Leftrightarrow7\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)\le6\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)+2017\)\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\le2017\)

Áp dụng BĐT Cauchy-Schwarz ta có:

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

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

\(\le\dfrac{1}{2a+b}+\dfrac{1}{2b+c}+\dfrac{1}{2c+a}\le\dfrac{1}{9}\left(\dfrac{2^2}{2a}+\dfrac{1^2}{b}\right)+\dfrac{1}{9}\left(\dfrac{2^2}{2b}+\dfrac{1^2}{c}\right)+\dfrac{1}{9}\left(\dfrac{2^2}{2c}+\dfrac{1^2}{a}\right)\)

\(\le\dfrac{1}{9}\left(\dfrac{3}{a}+\dfrac{3}{b}+\dfrac{3}{c}\right)\)\(=\dfrac{1}{3a}+\dfrac{1}{3b}+\dfrac{1}{3c}\le\sqrt{\left(\dfrac{1}{81}+\dfrac{1}{81}+\dfrac{1}{81}\right)\left(\dfrac{9}{a^2}+\dfrac{9}{b^2}+\dfrac{9}{c^2}\right)}\)

\(\le\sqrt{\dfrac{1}{81}\cdot3\cdot9\cdot2017}=\sqrt{\dfrac{2017}{3}}\)

Vậy \(T_{Max}=\sqrt{\dfrac{2017}{3}}\) khi \(a=b=c=\sqrt{\dfrac{3}{2017}}\)

So kimochiii~

Trước hết ta chứng minh bài toán quen thuộc:

Cho \(abc=1\) thì \(\frac{1}{ab+b+1}+\frac{1}{bc+c+1}+\frac{1}{ca+a+1}=1\)

\(VT=\frac{1}{ab+b+1}+\frac{1}{bc+c+abc}+\frac{b}{abc+ab+b}=\frac{1}{ab+b+1}+\frac{1}{c\left(b+1+ab\right)}+\frac{b}{1+ab+b}\)

\(=\frac{1}{ab+b+1}+\frac{ab}{b+1+ab}+\frac{b}{1+ab+b}=\frac{1+ab+b}{ab+b+1}=1\)

\(P=\sum\frac{1}{a^2+2b^2+3}=\sum\frac{1}{a^2+b^2+b^2+1+2}\le\sum\frac{1}{2ab+2b+2}=\frac{1}{2}\sum\frac{1}{ab+b+1}=\frac{1}{2}\)

\(\Rightarrow P_{max}=\frac{1}{2}\) khi \(a=b=c=1\)

\(P=\sum\frac{1}{a^2+1+2\left(b^2+1\right)}\le\sum\frac{1}{2a+4b}=\frac{1}{2}\sum\frac{1}{a+b+b}\le\frac{1}{18}\sum\left(\frac{1}{a}+\frac{2}{b}\right)\)

\(\Rightarrow P\le\frac{1}{18}\left(\frac{3}{a}+\frac{3}{b}+\frac{3}{c}\right)=\frac{1}{6}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\le\frac{1}{6}.3\sqrt[3]{\frac{1}{abc}}=\frac{1}{2}\)

\(\Rightarrow P_{max}=\frac{1}{2}\) khi \(a=b=c=1\)

Ta có: \(ab+bc+ac=abc+a+b+c\)

\(\Leftrightarrow ab-abc+bc-b+ac-a-c=0\)

\(\Leftrightarrow ab-abc+bc-b+ac-a+1-c=1\)

\(\Leftrightarrow ab\left(1-c\right)+b\left(c-1\right)+a\left(c-1\right)+\left(1-c\right)=1\)

\(\Leftrightarrow ab\left(1-c\right)-b\left(1-c\right)-a\left(1-c\right)+\left(1-c\right)=1\)

\(\Leftrightarrow\left(1-c\right)\left(ab-b-a+1\right)=1\)

\(\Leftrightarrow\left(1-a\right)\left(1-b\right)\left(1-c\right)=1\)

Ta có thể đặt x=1-a ; y=1-b; z=1-c => xyz=1

Nhưng trong đẳng thức cần chứng minh theo x;y;z

=> Thế: a=1-x; b=1-y; c=1-z vào được:

\(\frac{1}{3+ab-\left(2a+b\right)}=\frac{1}{3+\left(1-x\right)\left(1-y\right)-2\left(1-x\right)-\left(1-y\right)}=\frac{1}{1+x+xy}\)

Tương tự: \(\frac{1}{3+bc-\left(2b+c\right)}=\frac{1}{3+\left(1-y\right)\left(1-z\right)-2\left(1-y\right)-\left(1-z\right)}=\frac{1}{1+y+yz}\)

                  \(\frac{1}{3+ac-\left(2c+a\right)}=\frac{1}{3+\left(1-x\right)\left(1-z\right)-2\left(1-z\right)-\left(1-x\right)}=\frac{1}{1+z+zx}\)

Theo giả thiết xuz=1

=> \(VT=\frac{1}{1+x+xy}+\frac{1}{1+y+yz}+\frac{1}{1+z+zx}\)

             \(=\frac{1}{1+x+xy}+\frac{x}{x+xy+xyz}+\frac{xy}{xy+xyz+x^2yz}\)

            \(=\frac{1}{1+x+xy}+\frac{x}{x+xy+1}+\frac{xy}{xy+1+x}\)

            \(=\frac{1+x+xy}{1+x+xy}=1=VP\)

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