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\(\Leftrightarrow\frac{a}{bc}+\frac{b}{ac}+\frac{c}{ab}+2=\frac{1}{abc}\)
Đặt \(\left(\frac{a}{bc};\frac{b}{ac};\frac{c}{ab}\right)=\left(x;y;z\right)\)
\(\Rightarrow x+y+z+2=xyz\)
\(\Leftrightarrow\left(x+1\right)\left(y+1\right)+\left(y+1\right)\left(z+1\right)+\left(z+1\right)\left(x+1\right)=\left(x+1\right)\left(y+1\right)\left(z+1\right)\)
\(\Leftrightarrow\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}=1\)
\(\Leftrightarrow\frac{x}{x+1}+\frac{y}{y+1}+\frac{z}{z+1}=2\)
\(\Leftrightarrow\frac{a}{a+bc}+\frac{b}{c+ca}+\frac{c}{c+ab}=2\)
ÁP dụng BĐT cô-si, ta có \(a^3+b^3+c^3\ge3abc\Rightarrow\frac{a^3+b^3+c^3}{2abc}\ge\frac{3}{2}\)
Mà \(ab\le\frac{a^2+b^2}{2}\Rightarrow\frac{a^2+b^2}{c^2+ab}\ge\frac{2\left(a^2+b^2\right)}{2c^2+a^2+b^2}\)
Tương tự, ta có
\(\frac{a^2+b^2}{c^2+ab}+\frac{b^2+c^2}{a^2+bc}+\frac{c^2+a^2}{b^2+ac}\ge2\left(\frac{a^2+b^2}{a^2+c^2+b^2+c^2}+...\right)\)
Đặt \(\left(a^2+b^2;...\right)=\left(x;y;z\right)\)
Ta có VT\(\ge\frac{3}{2}+2\left(\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}\right)=\frac{3}{2}+2\left(\frac{x^2}{xy+zx}+\frac{y^2}{ỹ+yz}+\frac{z^2}{zx+zy}\right)\)
=> \(VT\ge\frac{3}{2}+2.\frac{\left(x+y+z\right)^2}{2\left(xy+yz+zx\right)}\ge\frac{3}{2}+3=\frac{9}{2}\)
=> \(A\ge\frac{9}{2}\left(ĐPCM\right)\)
Dấu = xảy ra <=> a=b=c>0
Giả sử b= min {a,b,c}
\(VT\ge\frac{a^3+b^3+c^3}{\frac{2\left(a+b+c\right)^3}{27}}+\frac{1}{2}\left(\Sigma\frac{\left(a+b\right)^2}{ab+c^2}+\Sigma\frac{\left(a-b\right)^2}{ab+c^2}\right)\)
\(\ge\left[\frac{27\left(a^3+b^3+c^3\right)}{2\left(a+b+c\right)^3}+\frac{2\left(a+b+c\right)^2}{\left(ab+bc+ca+a^2+b^2+c^2\right)}\right]\)
Sau khi quy đồng ta cần chứng minh biểu thức sau đây không âm:
Đó là điều hiển nhiên vì b = min {a,b,c}
Áp dụng BĐT cô-si, ta có \(a^3+b^3+c^3\ge3abc\Rightarrow\frac{a^3+b^3+c^3}{2abc}\ge\frac{3}{2}\)
Mà \(\frac{a^2+b^2}{c^2+ab}\ge\frac{a^2+b^2}{c^2+\frac{a^2+b^2}{2}}=2\frac{a^2+b^2}{2c^2+a^2+b^2}\)
tương tự thì \(P\ge\frac{3}{2}+2\left(\frac{a^2+b^2}{2c^2+a^2+b^2}+\frac{b^2+c^2}{2a^2+b^2+c^2}+\frac{c^2+a^2}{2b^2+a^2+c^2}\right)\)
Đặt \(\hept{\begin{cases}a^2+b^2=x\\b^2+c^2=y\\c^2+a^2=z\end{cases}}\)
ta có \(P\ge\frac{3}{2}+2\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\right)=\frac{3}{2}+2\left(\frac{x^2}{xy+xz}+\frac{y^2}{yz+yx}+\frac{z^2}{zx+zy}\right)\)
=>\(P\ge\frac{3}{2}+2.\frac{\left(x+y+z\right)^2}{2\left(xy+yz+zx\right)}\ge\frac{3}{2}+2.\frac{3\left(xy+yz+zx\right)}{2\left(xy+yz+zx\right)}\ge\frac{3}{2}+3=\frac{9}{2}\)
dấu xảy ra <>a=b=c>0
Vậy ...
^_^
1. Vai trò a, b, c như nhau. Không mất tính tổng quát. Giả sử \(a\ge b\ge0\)
Mà \(ab+bc+ca=3\). Do đó \(ab\ge1\)
Ta cần chứng minh rằng \(\frac{1}{1+a^2}+\frac{1}{1+b^2}\ge\frac{2}{1+ab}\left(1\right)\)
Và \(\frac{2}{1+ab}+\frac{1}{1+c^2}\ge\frac{3}{2}\left(2\right)\)
Thật vậy: \(\left(1\right)\Leftrightarrow\frac{1}{1+a^2}-\frac{1}{1+ab}+\frac{1}{1+b^2}-\frac{1}{1+ab}\ge0\\ \Leftrightarrow\left(ab-a^2\right)\left(1+b^2\right)+\left(ab-b^2\right)\left(1+a^2\right)\ge0\\ \Leftrightarrow\left(a-b\right)\left[-a\left(1+b^2\right)+b\left(1+a^2\right)\right]\ge0\\ \Leftrightarrow\left(a-b\right)^2\left(ab-1\right)\ge0\left(BĐT:đúng\right)\)
\(\left(2\right)\Leftrightarrow c^2+3-ab\ge3abc^2\\ \Leftrightarrow c^2+ca+bc\ge3abc^2\Leftrightarrow a+b+c\ge3abc\)
BĐT đúng, vì \(\left(a+b+c\right)^2>3\left(ab+bc+ca\right)=q\)
và \(ab+bc+ca\ge3\sqrt[3]{\left(abc\right)^2}\)
Nên \(a+b+c\ge3\ge3abc\)
Từ (1) và (2) ta có \(\frac{1}{1+a^2}+\frac{1}{1+b^2}+\frac{1}{1+c^2}\ge\frac{3}{2}\)
Dấu ''='' xảy ra \(\Leftrightarrow a=b=c=1\)
Áp dụng BĐT Cauchy dạng \(\frac{9}{x+y+z}\le\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\), ta được
\(\frac{9}{a+3b+2c}=\frac{1}{a+c+b+c+2b}\le\frac{1}{9}\left(\frac{1}{a+c}+\frac{1}{b+c}+\frac{1}{2b}\right)\)
Do đó ta được
\(\frac{ab}{a+3b+2c}\le\frac{ab}{9}\left(\frac{1}{a+c}+\frac{1}{b+c}+\frac{1}{2b}\right)=\frac{1}{9}\left(\frac{ab}{a+c}+\frac{ab}{b+c}+\frac{a}{2}\right)\)
Hoàn toàn tương tự ta được
\(\frac{bc}{2a+b+3c}\le\frac{1}{9}\left(\frac{bc}{a+b}+\frac{bc}{b+c}+\frac{b}{2}\right);\frac{ac}{3a+2b+c}\le\frac{1}{9}\left(\frac{ac}{a+b}+\frac{ac}{b+c}+\frac{c}{2}\right)\)
Cộng theo vế các BĐT trên ta được
\(\frac{ab}{a+3b+2c}+\frac{bc}{b+3c+2a}+\frac{ca}{c+3a+2b}\le\frac{1}{9}\left(\frac{ac+bc}{a+b}+\frac{ab+ac}{b+c}+\frac{bc+ab}{a+c}+\frac{a+b+c}{2}\right)=\frac{a+b+c}{6}\)Vậy BĐT đc CM
ĐẲng thức xảy ra khi và chỉ khi a = b = c >0
a)\(VT=\sum_{cyc}\frac{ab^3+ab^2c+a^2bc}{\left(a^2+bc+ca\right)\left(b^2+bc+ca\right)}\le\frac{\sum_{cyc}\left(ab^3+ab^2c+a^2bc\right)}{\left(ab+bc+ca\right)^2}\)
\(=\frac{ab^3+bc^3+ca^3+2a^2bc+2ab^2c+2abc^2}{\left(ab+bc+ca\right)^2}\)\(\le\frac{\sum_{cyc}ab\left(a^2+b^2\right)+abc\left(a+b+c\right)}{\left(ab+bc+ca\right)^2}\)
\(=\frac{\left(ab+bc+ca\right)\left(a^2+b^2+c^2\right)}{\left(ab+bc+ca\right)^2}=\frac{a^2+b^2+c^2}{ab+bc+ca}=VP\)
\(\Leftrightarrow\frac{a}{bc}+\frac{b}{ac}+\frac{c}{ab}+2=\frac{1}{abc}\)
Đặt : \(\left(\frac{a}{bc};\frac{b}{ac};\frac{c}{ab}\right)=\left(x,y,z\right)\)
\(x+y+z+2=xyz\)
\(\Leftrightarrow\left(x+1\right)\left(y+1\right)+\left(y+1\right)\left(z+1\right)+\left(z+1\right)\left(x+1\right)\)
\(=\left(x+1\right)\left(y+1\right)\left(z+1\right)\)
\(\Leftrightarrow\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}+1=1\)
\(\frac{x}{x+1}+\frac{y}{y+1}+\frac{z}{z+1}=2\)
\(\Leftrightarrow\frac{a}{a+bc}+\frac{b}{b+ca}+\frac{c}{c+ab}=2\)