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\(VT\ge\dfrac{1}{\left(a^2+1\right)-1}+\dfrac{1}{\left(b^2+1\right)-1}+\dfrac{1}{\left(c^2+1\right)-1}+4-\dfrac{4}{ab+1}+4-\dfrac{4}{bc+1}+4-\dfrac{4}{ca+1}\)
\(VT\ge\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}-\dfrac{4}{ab+1}-\dfrac{4}{bc+1}-\dfrac{4}{ca+1}+12\)
Mặt khác \(a;b;c\ge1\Rightarrow\left(a-1\right)\left(b-1\right)\ge0\Rightarrow ab+1\ge a+b\) (và tương tự...)
\(\Rightarrow VT\ge\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}-\dfrac{4}{a+b}-\dfrac{4}{b+c}-\dfrac{4}{c+a}+12\)
\(VT\ge\dfrac{4}{\left(a+b\right)^2}+\dfrac{4}{\left(b+c\right)^2}+\dfrac{4}{\left(c+a\right)^2}-\dfrac{4}{a+b}-\dfrac{4}{b+c}-\dfrac{4}{c+a}+1+1+1+9\)
\(VT\ge\left(\dfrac{2}{a+b}-1\right)^2+\left(\dfrac{2}{b+c}-1\right)^2+\left(\dfrac{2}{c+a}-1\right)^2+9\ge9\)
\(\dfrac{ab}{a+3b+2c}=\dfrac{ab}{\left(a+c\right)+\left(b+c\right)+2b}\le\dfrac{1}{9}\left(\dfrac{ab}{a+c}+\dfrac{ab}{b+c}+\dfrac{ab}{2b}\right)\)
\(=\dfrac{1}{9}\left(\dfrac{ab}{a+c}+\dfrac{ab}{b+c}+\dfrac{a}{2}\right)\)
Tương tự:
\(\dfrac{bc}{b+3c+2a}\le\dfrac{1}{9}\left(\dfrac{bc}{a+b}+\dfrac{bc}{a+c}+\dfrac{b}{2}\right)\)
\(\dfrac{ac}{c+3a+2b}\le\dfrac{1}{9}\left(\dfrac{ac}{b+c}+\dfrac{ac}{a+b}+\dfrac{c}{2}\right)\)
Cộng vế:
\(P\le\dfrac{1}{9}\left(\dfrac{bc+ac}{a+b}+\dfrac{bc+ab}{a+c}+\dfrac{ab+ac}{b+c}+\dfrac{a+b+c}{2}\right)\)
\(P\le\dfrac{1}{9}.\left(a+b+c+\dfrac{a+b+c}{2}\right)=\dfrac{1}{2}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
a) BĐT cần cm tương đương ;
\(a-\dfrac{ab^2}{1+b^2}+b-\dfrac{bc^2}{1+c^2}+a-\dfrac{a^2c}{1+a^2}\ge\dfrac{3}{2}\)
\(\Leftrightarrow3-\left(\dfrac{ab^2}{1+b^2}+\dfrac{bc^2}{1+c^2}+\dfrac{ac^2}{1+c^2}\right)\ge\dfrac{3}{2}\)
\(\Leftrightarrow\left(\dfrac{ab^2}{1+b^2}+\dfrac{bc^2}{1+c^2}+\dfrac{ac^2}{1+c^2}\right)\le\dfrac{3}{2}\)
Áp dụng BĐT Cauchy
\(\Rightarrow\dfrac{ab^2}{1+b^2}\le\dfrac{ab^2}{2b}=\dfrac{ab}{2}\)
tương tự rồi cộng vế theo vế các BĐT lại
\(\Leftrightarrow\dfrac{ab^2}{1+b^2}+\dfrac{bc^2}{1+c^2}+\dfrac{ac^2}{1+c^2}\le\dfrac{ab+bc+ac}{2}\)
mặt khác \(ab+bc+ac\le\dfrac{\left(a+b+c\right)^2}{3}=3\)
\(\Rightarrow\dfrac{ab^2}{1+b^2}+\dfrac{bc^2}{1+c^2}+\dfrac{ac^2}{1+c^2}\le\dfrac{3}{2}\)
ĐPCM
Bài 1:
Ta có:
\(\text{VT}=\frac{a^2}{a+2b^2}+\frac{b^2}{b+2c^2}+\frac{c^2}{c+2a^2}\)
\(=a-\frac{2ab^2}{a+2b^2}+b-\frac{2bc^2}{b+2c^2}+c-\frac{2ca^2}{c+2a^2}=(a+b+c)-2\left(\frac{ab^2}{a+2b^2}+\frac{bc^2}{b+2c^2}+\frac{ca^2}{c+2a^2}\right)\)
\(=3-2M(*)\)
Áp dụng BĐT Cauchy ta có:
\(M=\frac{ab^2}{a+b^2+b^2}+\frac{bc^2}{b+c^2+c^2}+\frac{ca^2}{c+a^2+a^2}\leq \frac{ab^2}{3\sqrt[3]{ab^4}}+\frac{bc^2}{3\sqrt[3]{bc^4}}+\frac{ca^2}{3\sqrt[3]{ca^4}}\)
\(\Leftrightarrow M\leq \frac{1}{3}(\sqrt[3]{a^2b^2}+\sqrt[3]{b^2c^2}+\sqrt[3]{c^2a^2})\)
Tiếp tục áp dụng BĐT Cauchy:
\(\sqrt[3]{a^2b^2}+\sqrt[3]{b^2c^2}+\sqrt[3]{c^2a^2}\leq \frac{ab+ab+1}{3}+\frac{bc+bc+1}{3}+\frac{ca+ca+1}{3}=\frac{2(ab+bc+ac)+3}{3}\)
Mà \(ab+bc+ac\leq \frac{(a+b+c)^2}{3}=3\) (quen thuộc)
\(\Rightarrow M\leq \frac{1}{3}.\frac{2.3+3}{3}=1(**)\)
Từ \((*);(**)\Rightarrow \text{VT}\geq 3-2.1=1\)
(đpcm)
Dấu bằng xảy ra khi $a=b=c=1$
Bài 2:
Áp dụng BĐT Cauchy -Schwarz:
\(\text{VT}=\frac{a^3}{a^2+a^2b^2}+\frac{b^3}{b^2+b^2c^2}+\frac{c^3}{c^2+a^2c^2}\geq \frac{(a\sqrt{a}+b\sqrt{b}+c\sqrt{c})^2}{a^2+a^2b^2+b^2+b^2c^2+c^2+c^2a^2}\)
hay:
\(\text{VT}\geq \frac{(a\sqrt{a}+b\sqrt{b}+c\sqrt{c})^2}{1+a^2b^2+b^2c^2+c^2a^2}(*)\)
Mặt khác, theo BĐT Cauchy ta dễ thấy:
\(a^4+b^4+c^4\geq a^2b^2+b^2c^2+c^2a^2\)
\(\Rightarrow (a^2+b^2+c^2)^2\geq 3(a^2b^2+b^2c^2+c^2a^2)\)
\(\Leftrightarrow 1\geq 3(a^2b^2+b^2c^2+c^2a^2)\Rightarrow a^2b^2+b^2c^2+c^2a^2\leq \frac{1}{3}(**)\)
Từ \((*);(**)\Rightarrow \text{VT}\geq \frac{(a\sqrt{a}+b\sqrt{b}+c\sqrt{c})^2}{1+\frac{1}{3}}=\frac{3}{4}(a\sqrt{a}+b\sqrt{b}+c\sqrt{c})^2\)
Ta có đpcm
Dấu bằng xảy ra khi \(a=b=c=\frac{1}{\sqrt{3}}\)
a) ta có
\(3\left(a+b+c\right)=\left(a^2+b^2+c^2\right)\left(a+b+c\right)\)
\(=a^3+b^3+c^3+a^2b+ab^2+b^2c+bc^2+a^2c+ac^2\)
\(=\left(a^3+ab^2\right)+\left(b^3+bc^2\right)+\left(c^3+ca^2\right)+a^2b+b^2c+c^2a\)
Áp dụng BĐT Cauchy ta có
\(a^3+ab^2\ge2a^2b\) ; \(b^3+bc^2\ge2b^2c\) ; \(c^3+ca^2\ge2c^2a\)
\(\left(a^3+ab^2\right)+\left(b^3+bc^2\right)+\left(c^3+ca^2\right)+a^2b+b^2c+c^2a\ge3\left(a^2b+b^2c+c^2a\right)\)\(\Rightarrow3\left(a+b+c\right)\ge3\left(a^2b+b^2c+c^2a\right)\)
\(\Rightarrow a+b+c\ge a^2b+b^2c+c^2a\) (1)
Áp dụng BĐT C.B.S ta có
\(\left(a+b+c\right)^2\le3\left(a^2+b^2+c^2\right)=9\)
\(\Rightarrow a+b+c\le3\) (2)
từ (1) và (2) ta được đpcm
b) Áp dụng BĐT Cauchy ta có :
\(ab\le\dfrac{a^2+b^2}{2}=\dfrac{3-c^2}{2}\) tương tự
\(bc\le\dfrac{3-a^2}{2}\) ; \(ac\le\dfrac{3-b^2}{2}\)
BĐT cần chứng minh trở thành :
\(\dfrac{3-a^2}{2\left(3+a^2\right)}+\dfrac{3-b^2}{2\left(3+b^2\right)}+\dfrac{3-c^2}{2\left(3+c^2\right)}\le\dfrac{3}{4}\)
Ta chứng minh BĐT phụ sau
\(\dfrac{3-c^2}{2\left(3+c^2\right)}\le\dfrac{c^2}{4}\)\(\Leftrightarrow12-4c^2\le2c^2\left(3+c^2\right)\Leftrightarrow c^4+5c^2+6\ge0\)
\(\Leftrightarrow\left(c^2+2\right)\left(c^2+3\right)\ge0\) (luôn đúng)
tương tự : \(\dfrac{3-a^2}{2\left(3+c^2\right)}\le\dfrac{a^2}{4}\) ; \(\dfrac{3-b^2}{2\left(3+b^2\right)}\le\dfrac{b^2}{4}\)
Cộng Ba vế BĐT trên lại ta có:
\(\dfrac{3-a^2}{2\left(3+a^2\right)}+\dfrac{3-b^2}{2\left(3+b^2\right)}+\dfrac{3-c^2}{2\left(3+c^2\right)}\le\dfrac{a^2+b^2+c^2}{4}=\dfrac{3}{4}\)
Vậy ta có đpcm
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{b^2c}{a^3\left(b+c\right)}+\dfrac{b+c}{4bc}+\dfrac{1}{2b}\ge3\sqrt[3]{\dfrac{b^2c\left(b+c\right)}{8a^3\left(b+c\right)b^2c}}=\dfrac{3}{2a}\\\dfrac{c^2a}{b^3\left(c+a\right)}+\dfrac{c+a}{4ca}+\dfrac{1}{2c}\ge3\sqrt[3]{\dfrac{c^2a\left(c+a\right)}{8b^3\left(c+a\right)c^2a}}=\dfrac{3}{2b}\\\dfrac{a^2b}{c^3\left(a+b\right)}+\dfrac{a+b}{4ab}+\dfrac{1}{2a}\ge3\sqrt[3]{\dfrac{a^2b\left(a+b\right)}{8c^3\left(a+b\right)a^2b}}=\dfrac{3}{2c}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{b^2c}{a^3\left(b+c\right)}+\dfrac{1}{4c}+\dfrac{1}{4b}+\dfrac{1}{2b}\ge\dfrac{3}{2a}\\\dfrac{c^2a}{b^3\left(c+a\right)}+\dfrac{1}{4a}+\dfrac{1}{4c}+\dfrac{1}{2c}\ge\dfrac{3}{2b}\\\dfrac{a^2b}{c^3\left(a+b\right)}+\dfrac{1}{4b}+\dfrac{1}{4a}+\dfrac{1}{2a}\ge\dfrac{3}{2c}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{b^2c}{a^3\left(b+c\right)}+\dfrac{1}{4c}+\dfrac{3}{4b}\ge\dfrac{3}{2a}\\\dfrac{c^2a}{b^3\left(c+a\right)}+\dfrac{1}{4a}+\dfrac{3}{4c}\ge\dfrac{3}{2b}\\\dfrac{a^2b}{c^3\left(a+b\right)}+\dfrac{1}{4b}+\dfrac{3}{4a}\ge\dfrac{3}{2c}\end{matrix}\right.\)
\(\Rightarrow VT+\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)+\dfrac{3}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge\dfrac{3}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
\(\Rightarrow VT+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{3}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
\(\Rightarrow VT\ge\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
\(\Leftrightarrow\dfrac{b^2c}{a^3\left(b+c\right)}+\dfrac{c^2a}{b^3\left(c+a\right)}+\dfrac{a^2b}{c^3\left(a+b\right)}\ge\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\) ( đpcm )
Từ \(a^2b^2+b^2c^2+c^2a^2\ge a^2b^2c^2\)\(\Rightarrow\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}=1\)
bài này tui làm rồi ở đây
\(\sum\dfrac{a}{\left(a^2+1\right)+2b+2}\le\sum\dfrac{a}{2\left(a+b+1\right)}=\dfrac{1}{2}\)
Ta có: \(\dfrac{a^3}{a^2+2b^2}=a-\dfrac{2ab^2}{a^2+2b^2}\ge a-\dfrac{2ab^2}{3\sqrt[3]{a^2b^4}}=a-\dfrac{2}{3}\sqrt[3]{ab^2}\ge a-\dfrac{2}{9}\left(a+b+b\right)=a-\dfrac{2}{9}\left(a+2b\right)\) Chứng minh tương tự ta được:
\(\dfrac{b^3}{b^2+2c^2}\ge b-\dfrac{2}{9}\left(b+2c\right);\dfrac{c^3}{c^2+2a^2}\ge c-\dfrac{2}{9}\left(c+2a\right)\)
\(\Rightarrow\dfrac{a^3}{a^2+2b^2}+\dfrac{b^3}{b^2+2c^2}+\dfrac{c^3}{c^2+2a^2}\ge a+b+c-\dfrac{2}{9}\left(a+2b+b+2c+c+2a\right)=a+b+c-\dfrac{2}{9}\left(3a+3b+3c\right)=\dfrac{1}{3}\left(a+b+c\right)\ge\dfrac{1}{3}\cdot3\sqrt[3]{abc}=1\)Dấu = xảy ra \(\Leftrightarrow a=b=c=1\)