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\(Q=\frac{a}{b+2020-a}+\frac{b}{c+2020-b}+\frac{c}{a+2020-c}\)
\(Q=\frac{a}{b+a+b+c-a}+\frac{b}{c+a+b+c-b}+\frac{c}{a+a+b+c-c}\)
\(Q=\frac{a}{2b+c}+\frac{b}{2c+a}+\frac{c}{2a+b}\)
Áp dụng BĐT Cauchy-Schwarz:
\(Q=\frac{a^2}{a\cdot\left(2b+c\right)}+\frac{b^2}{b\cdot\left(2c+a\right)}+\frac{c^2}{c\cdot\left(2a+b\right)}\ge\frac{\left(a+b+c\right)^2}{3\cdot\left(ab+bc+ca\right)}\ge\frac{3\cdot\left(ab+bc+ca\right)}{3\cdot\left(ab+bc+ca\right)}=1\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=\frac{2020}{3}\)
\(M=\sqrt{\frac{\left(a^2+2020\right)\left(b^2+2020\right)}{c^2+2020}}\)
\(=\sqrt{\frac{\left(a^2+ab+bc+ac\right)\left(b^2+ab+bc+ac\right)}{c^2+ab+bc+ac}}\)
\(=\sqrt{\frac{\left(a+b\right)\left(a+c\right)\left(b+c\right)\left(b+a\right)}{\left(c+a\right)\left(c+b\right)}}\)
\(=a+b\) là 1 số hữu tỉ
=> M là 1 số hữu tỉ (đpcm)
Đặt \(\left(b+c,c+a,a+b\right)\rightarrow\left(x,y,z\right)\)thì \(x,y,z>0\)và \(a=\frac{y+z-x}{2};b=\frac{z+x-y}{2};c=\frac{x+y-z}{2}\)
Bất đẳng thức cần chứng minh trở thành: \(\frac{y+z-x}{2x}+\frac{25\left(z+x-y\right)}{2y}+\frac{4\left(x+y-z\right)}{2z}>2\)
Xét \(VT=\left(\frac{y}{2x}+\frac{z}{2x}-\frac{1}{2}\right)+\left(\frac{25z}{2y}+\frac{25x}{2y}-\frac{25}{2}\right)+\left(\frac{2x}{z}+\frac{2y}{z}-2\right)\)\(=\left(\frac{y}{2x}+\frac{25x}{2y}\right)+\left(\frac{25z}{2y}+\frac{2y}{z}\right)+\left(\frac{z}{2x}+\frac{2x}{z}\right)-15\)\(\ge2\sqrt{\frac{y}{2x}.\frac{25x}{2y}}+2\sqrt{\frac{25z}{2y}.\frac{2y}{z}}+2\sqrt{\frac{z}{2x}.\frac{2x}{z}}-15=2\)(BĐT Cauchy)
Đẳng thức xảy ra khi \(10x=2y=5z\)hay \(10\left(b+c\right)=2\left(c+a\right)=5\left(a+b\right)\)\(\Rightarrow\hept{\begin{cases}10b+8c=2a\\5b+10c=5a\end{cases}}\Leftrightarrow\hept{\begin{cases}2a=10b+8c\\2a=2b+4c\end{cases}}\Leftrightarrow8b+4c=0\)(Vô lí vì 8b + 4c > 0 với mọi b,c dương)
Vậy dấu bằng không xảy ra
\(\sqrt{a\left(a+b+c\right)+bc}=\sqrt{\left(a+b\right)\left(c+a\right)}\ge\sqrt{\left(\sqrt{ac}+\sqrt{ab}\right)^2}=\sqrt{ac}+\sqrt{ab}\)
\(\Rightarrow\frac{a}{a+\sqrt{2020a+bc}}\le\frac{a}{a+\sqrt{ac}+\sqrt{ab}}=\frac{\sqrt{a}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}\)
Tương tự: \(\frac{b}{b+\sqrt{2020b+ca}}\le\frac{\sqrt{b}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}\) ; \(\frac{c}{c+\sqrt{2020c+ab}}\le\frac{\sqrt{c}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}\)
Cộng vế với vế: \(P\le1\)
Dấu "=" xảy ra khi \(a=b=c=...\)
\(\Rightarrow5a^2+\frac{5}{2}\left(b+c\right)^2\le9a\left(b+c\right)+18bc\le9a\left(b+c\right)+\frac{9}{2}\left(b+c\right)^2\)
\(\Rightarrow5a^2\le9a\left(b+c\right)+2\left(b+c\right)^2\)
\(\Rightarrow5a^2\le18.\frac{a}{2}\left(b+c\right)+2\left(b+c\right)^2\le9\left(\frac{a^2}{4}+\left(b+c\right)^2\right)+2\left(b+c\right)^2\)
\(\Rightarrow\frac{11a^2}{4}\le11\left(b+c\right)^2\Rightarrow b+c\ge\frac{a}{2}\Rightarrow\frac{b+c}{a}\ge\frac{1}{2}\)
\(P\ge1010\left(\frac{b+c}{a}\right)^2+\frac{a}{b+c}\)
Đặt \(\frac{b+c}{a}=x\ge\frac{1}{2}\Rightarrow P\ge1010x^2+\frac{1}{x}=1006x^2+4x^2+\frac{1}{2x}+\frac{1}{2x}\)
\(\Rightarrow P\ge1006.\left(\frac{1}{2}\right)^2+3\sqrt[3]{\frac{4x^2}{4x^2}}=...\)
Ta có: \(ab+bc+ca=abc\)
\(\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1\)
Đặt: \(A=\frac{a}{bc\left(a+1\right)}+\frac{b}{ca\left(b+1\right)}+\frac{c}{ab\left(c+1\right)}\)
\(\Rightarrow A=\frac{\frac{1}{b}.\frac{1}{c}}{1+\frac{1}{a}}+\frac{\frac{1}{c}.\frac{1}{a}}{1+\frac{1}{b}}+\frac{\frac{1}{b}.\frac{1}{a}}{1+\frac{1}{c}}\)
Đặt: \(\hept{\begin{cases}x=\frac{1}{a}\\y=\frac{1}{b}\\z=\frac{1}{c}\end{cases}}\Rightarrow x+y+z=1\)
\(A=\frac{xy}{z+1}+\frac{yz}{x+1}+\frac{zx}{y+1}\)
Ta có: \(\frac{xy}{z+1}=\frac{xy}{\left(z+x\right)+\left(z+y\right)}\le\frac{1}{4}\left(\frac{xy}{x+z}+\frac{xy}{y+z}\right)\)
Chứng minh tương tự ta được:
\(\frac{yz}{x+1}\le\frac{yz}{x+y}+\frac{yz}{x+z}\)
\(\frac{zx}{y+1}\le\frac{zx}{x+y}+\frac{zx}{y+z}\)
Cộng vế với vế:
\(\Rightarrow A\le\frac{1}{4}\left(x+y+z\right)=\frac{1}{4}\left(đpcm\right)\)
