\(\left(a+b\right)\left(b+c\right)\left(c+a\right)+abc\)

\(=abc+a^2b+ab^2+a^2c+ac^2+b^2c+bc^2+abc+abc\)

\(=\left(a+b+c\right)\left(ab+bc+ca\right)\)( phân tích nhân tử các kiểu )

\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge\left(a+b+c\right)\left(ab+bc+ca\right)-abc\left(1\right)\)

\(a+b+c\ge3\sqrt[3]{abc};ab+bc+ca\ge3\sqrt[3]{a^2b^2c^2}\Rightarrow\left(a+b+c\right)\left(ab+bc+ca\right)\ge9abc\)

\(\Rightarrow-abc\ge\frac{-\left(a+b+c\right)\left(ab+bc+ca\right)}{9}\)

Khi đó:\(\left(a+b+c\right)\left(ab+bc+ca\right)-abc\)

\(\ge\left(a+b+c\right)\left(ab+bc+ca\right)-\frac{\left(a+b+c\right)\left(ab+bc+ca\right)}{9}\)

\(=\frac{8\left(a+b+c\right)\left(ab+bc+ca\right)}{9}\left(2\right)\)

Từ ( 1 ) và ( 2 ) có đpcm

Ta có: \(\left(a+\frac{1}{a}\right)\left(b+\frac{1}{b}\right)\left(c+\frac{1}{c}\right)\)

\(=\left(ab+\frac{1}{ab}+\frac{a}{b}+\frac{b}{a}\right)\left(c+\frac{1}{c}\right)\)

\(=\left[ab+\frac{1}{16ab}+\frac{15}{16ab}+\left(\frac{a}{b}+\frac{b}{a}\right)\right]\left(c+\frac{1}{c}\right)\)

\(\ge\left[2\sqrt{ab.\frac{1}{16ab}}+\frac{15}{4\left(a+b\right)^2}+2\sqrt{\frac{a}{b}.\frac{b}{a}}\right]\left(2\sqrt{c.\frac{1}{c}}\right)\)

\(\ge\frac{25}{2}\left(Đpcm\right)\)

Dấu " = " xảy ra \(\Leftrightarrow a=b=\frac{1}{2};c=1\)

Ta có \(\left(a+b+1\right).\left(a^2+b^2\right)+\frac{4}{a+b}\)

\(\ge\left(a+b+1\right).2ab+\frac{4}{a+b}\)

\(=2.\left(a+b\right)+2+\frac{4}{a+b}\)

\(=a+b+2+a+b+\frac{4}{a+b}\)

\(\ge2.\sqrt{a.b}+2+2.\sqrt{\left(a+b\right).\frac{4}{a+b}}=2+2+2\sqrt{4}\)

\(=2+2+4=8\)

Vậy\(\left(a+b+1\right).\left(a^2+b^2\right)+\frac{4}{a+b}\ge8\)với ab=1

1)

   +)  Ta có

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

       \(\Leftrightarrow a^2+b^2-2ab\ge0\)

        \(\Leftrightarrow a^2+b^2\ge2ab\)

        \(\Leftrightarrow2\left(a^2+b^2\right)\ge a^2+b^2+2ab\)

        \(\Leftrightarrow2\left(a^2+b^2\right)\ge\left(a+b\right)^2\)

        \(\Leftrightarrow a^2+b^2\ge\frac{1}{2}\left(a+b\right)^2\)  ( đpcm )

     + )   Theo phần trên

             \(a^2+b^2\ge2ab\)

           \(\Leftrightarrow a^2+b^2+2ab\ge4ab\)

           \(\Leftrightarrow\left(a+b\right)^2\ge4ab\)

            \(\Leftrightarrow ab\le\frac{1}{4}\left(a+b\right)^2\)  ( đpcm )

2, 

Ta có: \(5\left(x^2+y^2+z^2\right)-9x\left(y+z\right)-18yz=0\Leftrightarrow5x^2-9x\left(y+z\right)+5\left(y+z\right)^2=28yz\le7\left(y+z\right)^2\)\(\Leftrightarrow5x^2-9x\left(y+z\right)-2\left(y+z\right)^2\le0\Leftrightarrow5\left(\frac{x}{y+z}\right)^2-9.\frac{x}{y+z}-2\le0\)\(\Leftrightarrow\left(5.\frac{x}{y+z}+1\right)\left(\frac{x}{y+z}-2\right)\le0\Leftrightarrow\frac{x}{y+z}\le2\)(Do \(5.\frac{x}{y+z}+1>0\forall x,y,z>0\))

\(\Rightarrow E=\frac{2x-y-z}{y+z}=2.\frac{x}{y+z}-1\le2.2-1=3\)

Đẳng thức xảy ra khi \(y=z=\frac{x}{4}\)

Cho a, b, c khác 0 thoả mãn a+b+c=0. Tính $A=\left(1+\frac{a}{b}\right)+\left(1+\frac{b}{c}\right)+\left(1+\frac{c}{a}\right)$A=(1+ab )+(1+bc )+(1+ca )

Cho a, b, c khác 0 thoả mãn a+b+c=0. Tính $A=\left(1+\frac{a}{b}\right)+\left(1+\frac{b}{c}\right)+\left(1+\frac{c}{a}\right)$A=(1+ab )+(1+bc )+(1+ca )

Khó quá do anh thien

\(A=3+\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\)

$A=\left(1+\frac{a}{b}\right)+\left(1+\frac{b}{c}\right)+\left(1+\frac{c}{a}\right)$

$A=\left(1+\frac{a}{b}\right)+\left(1+\frac{b}{c}\right)+\left(1+\frac{c}{a}\right)$$A=\left(1+\frac{a}{b}\right)+\left(1+\frac{b}{c}\right)+\left(1+\frac{c}{a}\right)$$A=\left(1+\frac{a}{b}\right)+\left(1+\frac{b}{c}\right)+\left(1+\frac{c}{a}\right)$

Thiếu đề nha bạn 

Cho a, b, c khác 0 thoả mãn a+b+c=0. Tính 

Áp dụng BĐT Cosi ta có \(\frac{ab}{a^2+b^2}+\frac{a^2+b^2}{4ab}\ge2\sqrt{\frac{ab}{a^2+b^2}.\frac{a^2+b^2}{4ab}}=1\)

Tương tự \(\frac{bc}{b^2+c^2}+\frac{b^2+c^2}{4bc}\ge1\) \(\frac{ca}{c^2+a^2}+\frac{c^2+a^2}{4ca}\ge1\)

Khi đó BĐT sẽ được chứng minh nếu ta chỉ ra được

\(\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-\left(\frac{a^2+b^2}{4ab}+\frac{b^2+c^2}{4bc}+\frac{c^2+a^2}{4ca}\right)\ge\frac{3}{4}\)

\(\Leftrightarrow\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\left(\frac{a}{4b}+\frac{b}{4a}+\frac{b}{4c}+\frac{c}{4b}+\frac{a}{4c}+\frac{c}{4a}\right)\right)\ge\frac{3}{4}\)

\(\Leftrightarrow\frac{1}{4}\left(\frac{a+b+c}{a}+\frac{a+b+c}{b}+\frac{a+b+c}{c}-\frac{a+c}{b}-\frac{b+c}{a}-\frac{c+a}{b}\right)\ge\frac{3}{4}\)(do \(a+b+c=1\))

\(\Leftrightarrow\frac{3}{4}\ge\frac{3}{4}\) luôn đúng. Từ đó suy ba BĐT được chứng minh. Dấu "=" xảy ra khi \(a=b=c=\frac{1}{3}\)