Đề sai rồi

Đây nhé

.

\(\left(x^2+10x+8\right)^2=\left(8x+4\right)\left(x^2+8x+7\right)\)

\(\Leftrightarrow x^4+12x^3+48x^2+72x+36=0\)

\(\Leftrightarrow\left(x^4+16x^3+36x^2\right)+\left(12x^2+72x\right)+36=0\)

\(\Leftrightarrow\left(x^2+6x\right)^2+12\left(x^2+6x\right)+36=0\)

\(\Leftrightarrow\left(x^2+6x+6\right)^2=0\)

Làm nốt

Đặt \(a+b-c=x;b+c-a=y;c+a-b=z\)

\(\Rightarrow a=\frac{z+x}{2};b=\frac{x+y}{2};c=\frac{y+z}{2}\)

Bài toán cần chứng minh:

\(\frac{\left(x+y\right)\left(z+x\right)}{4x}+\frac{\left(x+y\right)\left(y+z\right)}{4y}+\frac{\left(y+z\right)\left(z+x\right)}{4z}\ge x+y+z\)

Ta có:

\(VT=\frac{3}{4}\left(x+y+z\right)+\frac{1}{4}\left(\frac{xy}{z}+\frac{yz}{x}+\frac{zx}{y}\right)\)

\(=\frac{3}{4}\left(x+y+z\right)+\frac{1}{4xyz}\left(x^2y^2+y^2z^2+z^2x^2\right)\)

\(\ge\frac{3}{4}\left(x+y+z\right)+\frac{1}{4xyz}\left(x+y+z\right)xyz\)

\(=x+y+z=VP\)

Ta có:

\(a^3+b^3+c^3=3abc\)

\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)

\(\Leftrightarrow a+b+c=0\)( Vì a,b,c đôi 1 khác nhau nên \(\left(a^2+b^2+c^2-ab-bc-ca\right)\ne0\)

Ta có:

\(P=\frac{ab^2}{a^2+b^2-c^2}+\frac{bc^2}{b^2+c^2-a^2}+\frac{ca^2}{c^2+a^2-b^2}\)

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

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

\(=\frac{b^2}{a+c-b}+\frac{c^2}{b+a-c}+\frac{a^2}{c+b-a}\)

\(=\frac{b^2}{-b-b}+\frac{c^2}{-c-c}+\frac{a^2}{-a-a}=-\frac{b}{2}-\frac{c}{2}-\frac{a}{2}=0\)

\(B=\frac{7y^2-4xy}{x^2-2xy+2y^2}=\frac{\left(-x^2+2xy-2y^2\right)+\left(x^2-6xy+9y^2\right)}{x^2-2xy+2y^2}=\frac{\left(x-3y\right)^2}{x^2-2xy+2y^2}-1\ge-1\)

Dấu =  xảy ra khi x = 3y

Ta có:

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

\(\ge\frac{4}{2a+4b+2c}+\frac{4}{2a+2b+4c}+\frac{4}{4a+2b+2c}\)

\(\ge\frac{2}{a+2b+c}+\frac{2}{a+b+2c}+\frac{2}{2a+b+c}\)

Vậy

\(\frac{1}{a+3b}+\frac{1}{b+3c}+\frac{1}{c+3a}\ge\frac{1}{2a+b+c}+\frac{1}{a+2b+c}+\frac{1}{a+b+2c}+\)

Đặt \(\hept{\begin{cases}a=3x^2-4x+1\\b=3x^2+2x+1\end{cases}}\)

\(\Rightarrow b-a=6x\)

Thế lại bài toán được

\(\frac{2}{a}+\frac{13}{b}=\frac{36}{b-a}\)

\(\Leftrightarrow2b\left(b-a\right)+13a\left(b-a\right)-36ab=0\)

\(\Leftrightarrow2b^2-25ab-13a^2=0\)

\(\Leftrightarrow\left(ab+2b^2\right)-\left(13a^2+26ab\right)=0\)

\(\Leftrightarrow\left(a+2b\right)\left(b-13a\right)=0\)

Làm nốt nha

accept đi mình ở An Quý nè

Xét \(x=0\Rightarrow B=0\left(1\right)\)

Xét \(x\ne0\) thì ta có:

\(\frac{1}{B}=\frac{x^2+20x+100}{x}=x+\frac{100}{x}+20\ge2\sqrt{x.\frac{100}{x}}+20=40\)

\(\Rightarrow B\le\frac{1}{40}\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\)GTLN là \(\frac{1}{40}\)

\(\frac{a}{-a+2b+2c}+\frac{b}{2a-b+2c}+\frac{2}{2a+2b-c}\)

\(=\frac{a^2}{-a^2+2ab+2ca}+\frac{b^2}{2ab-b^2+2bc}+\frac{c^2}{2ac+2bc-c^2}\)

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

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

\(\ge\frac{\left(a+b+c\right)^2}{\frac{6\left(a+b+c\right)^2}{3}-\left(a+b+c\right)^2}=1\)