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\(1+a^2=a^2+ab+bc+ca=\left(a+b\right)\left(c+a\right)\)
Tương tự, ta có: \(1+b^2=\left(a+b\right)\left(b+c\right)\)\(;\)\(1+c^2=\left(b+c\right)\left(c+a\right)\)
\(\Rightarrow\)\(\frac{2}{\sqrt{\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)}}=\frac{2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\) ( do a, b, c dương )
\(\frac{a}{1+a^2}+\frac{b}{1+b^2}+\frac{c}{1+c^2}=\frac{a\left(b+c\right)+b\left(c+a\right)+c\left(a+b\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=\frac{2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
...
\(\frac{\left(b+c\right)}{a}+\frac{\left(c+a\right)}{b}+\frac{\left(a+b\right)}{c}\)
\(=\frac{b}{a}+\frac{c}{a}+\frac{c}{b}+\frac{a}{b}+\frac{a}{c}+\frac{b}{c}\)
\(=\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\)
mà \(\frac{a}{b}+\frac{b}{a}\ge2\)(dễ chứng minh)
chứng minh tương tự ta có
\(\frac{\left(b+c\right)}{a}+\frac{\left(c+a\right)}{b}+\frac{\left(a+b\right)}{c}\)\(\ge\)6
\(\left(\frac{\left(b+c\right)}{a}+\frac{\left(c+a\right)}{b}+\frac{\left(a+b\right)}{c}\right)^2\ge6^2=36\)(2) (a>0; b>0; c>0)
tiếp theo chứng minh
\(36\ge4\left(ab+bc+ca\right)\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\)
\(18\ge2\left(ab+bc+ca\right)\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\)
\(18a^2+18b^2+18c^2\ge2ab+2bc+2ca\)
\(16\left(a^2+b^2+c^2\right)+\left(a^2-2ab+b^2\right)+\left(a^2-2ac+c^2\right)+\left(b^2-2bc+c^2\right)\ge0\)
\(16\left(a^2+b^2+c^2\right)+\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\) (bất đẳng thức luôn đúng )
suy ra bất đẳng thức
\(36\ge4\left(ab+bc+ca\right)\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\)luôn đúng (2)
từ (1) và (2) suy ra
\(\left(\frac{\left(b+c\right)}{a}+\frac{\left(c+a\right)}{b}+\frac{\left(a+b\right)}{c}\right)^2\ge\text{}\text{36}\ge\)\(4\left(ab+bc+ca\right)\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\)
1) Áp dụng bunhiacopxki ta được \(\sqrt{\left(2a^2+b^2\right)\left(2a^2+c^2\right)}\ge\sqrt{\left(2a^2+bc\right)^2}=2a^2+bc\), tương tự với các mẫu ta được vế trái \(\le\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ac}+\frac{c^2}{2c^2+ab}\le1< =>\)\(1-\frac{bc}{2a^2+bc}+1-\frac{ac}{2b^2+ac}+1-\frac{ab}{2c^2+ab}\le2< =>\)
\(\frac{bc}{2a^2+bc}+\frac{ac}{2b^2+ac}+\frac{ab}{2c^2+ab}\ge1\)<=> \(\frac{b^2c^2}{2a^2bc+b^2c^2}+\frac{a^2c^2}{2b^2ac+a^2c^2}+\frac{a^2b^2}{2c^2ab+a^2b^2}\ge1\) (1)
áp dụng (x2 +y2 +z2)(m2+n2+p2) \(\ge\left(xm+yn+zp\right)^2\)
(2a2bc +b2c2 + 2b2ac+a2c2 + 2c2ab+a2b2). VT\(\ge\left(bc+ca+ab\right)^2\) <=> (ab+bc+ca)2. VT \(\ge\left(ab+bc+ca\right)^2< =>VT\ge1\) ( vậy (1) đúng)
dấu '=' khi a=b=c
Vì a+b+c=1 nên \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{a+b+c}{a}+\frac{a+b+c}{b}+\frac{a+b+c}{c}\)
\(=3+\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{b}{c}+\frac{c}{a}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)=2+\frac{a^2+b^2}{ab}+\frac{b^2+c^2}{bc}+\frac{c^2+a^2}{ca}\)
Do đó
\(\frac{ab}{a^2+b^2}+\frac{bc}{b^2+c^2}+\frac{ca}{c^2+a^2}+\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=\left(\frac{ab}{a^2+b^2}+\frac{a^2+b^2}{ab}\right)+\left(\frac{bc}{b^2+c^2}+\frac{b^2+c^2}{bc}\right)+\left(\frac{ca}{a^2+c^2}+\frac{c^2+a^2}{ca}\right)+\frac{3}{4}\)
\(\ge2\sqrt{\frac{ab}{a^2+b^2}\cdot\frac{a^2+b^2}{ab}}+2\sqrt{\frac{bc}{c^2+b^2}\cdot\frac{c^2+b^2}{bc}}+2\sqrt{\frac{ca}{a^2+c^2}+\frac{c^2+a^2}{ca}}+\frac{3}{4}\)
\(=2\cdot\frac{1}{2}+2\cdot\frac{1}{2}+\frac{2}{3}=\frac{15}{4}\)
Dấu "=" xảy ra <=> \(a=b=c=\frac{1}{3}\)
Câu 2/
\(\frac{a^2+bc}{a^2\left(b+c\right)}+\frac{b^2+ca}{b^2\left(c+a\right)}+\frac{c^2+ab}{c^2\left(a+b\right)}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
\(\Leftrightarrow\frac{a^2+bc}{a^2\left(b+c\right)}-\frac{1}{a}+\frac{b^2+ca}{b^2\left(c+a\right)}-\frac{1}{b}+\frac{c^2+ab}{c^2\left(a+b\right)}-\frac{1}{c}\ge0\)
\(\Leftrightarrow\frac{\left(b-a\right)\left(c-a\right)}{a^2\left(b+c\right)}+\frac{\left(a-b\right)\left(c-b\right)}{b^2\left(c+a\right)}+\frac{\left(a-c\right)\left(b-c\right)}{c^2\left(a+b\right)}\ge0\)
\(\Leftrightarrow a^4b^4+b^4c^4+c^4a^4-a^4b^2c^2-a^2b^4c^2-a^2b^2c^4\ge0\)
\(\Leftrightarrow a^4b^4+b^4c^4+c^4a^4\ge a^4b^2c^2+a^2b^4c^2+a^2b^2c^4\left(1\right)\)
Ma ta có: \(\hept{\begin{cases}a^4b^4+b^4c^4\ge2a^2b^4c^2\left(2\right)\\b^4c^4+c^4a^4\ge2a^2b^2c^4\left(3\right)\\c^4a^4+a^4b^4\ge2a^4b^2c^2\left(4\right)\end{cases}}\)
Cộng (2), (3), (4) vế theo vế rồi rút gọn cho 2 ta được điều phải chứng minh là đúng.
PS: Nếu nghĩ được cách khác đơn giản hơn sẽ chép lên cho b sau. Tạm cách này đã.
Ta có \(ab+bc+ca\ge3\sqrt[3]{a^2b^2c^2}\)\(\Rightarrow3\sqrt[3]{a^2b^2c^2}\le3\Leftrightarrow abc\le1\)
\(\Rightarrow\)\(\frac{1}{1+a^2\left(b+c\right)}\le\frac{1}{abc+a^2\left(b+c\right)}\)\(=\frac{1}{a\left(ab+bc+ca\right)}=\frac{1}{3a}\)
\(CMTT\Rightarrow\frac{1}{1+b^2\left(c+a\right)}\le\frac{1}{3b}\)
\(\frac{1}{1+c^2\left(a+b\right)}\le\frac{1}{3c}\)
\(\Rightarrow VT\le\frac{1}{3a}+\frac{1}{3b}+\frac{1}{3c}\)\(=\frac{ab+bc+ca}{3abc}=\frac{1}{abc}\)