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Áp dụng bất đẳng thức Bunyakovsky
\(\Rightarrow\sqrt{\left(\dfrac{8}{a^2}+\dfrac{9b^2}{2}+\dfrac{c^2a^2}{4}\right)\left[\left(\sqrt{2}\right)^2+\left(3\sqrt{2}\right)^2+2^2\right]}\ge\left(\sqrt{\dfrac{4}{a}+9b+ca}\right)^2\)
\(\Leftrightarrow2\sqrt{6}\sqrt{\dfrac{8}{a^2}+\dfrac{9b^2}{2}+\dfrac{c^2a^2}{4}}\ge\dfrac{4}{a}+9b+ac\)
Tương tự ta có \(\left\{{}\begin{matrix}2\sqrt{6}\sqrt{\left(\dfrac{8}{b^2}+\dfrac{9c^2}{2}+\dfrac{a^2b^2}{4}\right)}\ge\dfrac{4}{b}+9c+ab\\2\sqrt{6}\sqrt{\left(\dfrac{8}{c^2}+\dfrac{9a^2}{2}+\dfrac{b^2c^2}{4}\right)}\ge\dfrac{4}{c}+9a+bc\end{matrix}\right.\)
\(\Rightarrow2\sqrt{6}S\ge\dfrac{4}{a}+9a+\dfrac{4}{b}+9b+\dfrac{4}{c}+9c+ab+bc+ac\)
\(\Leftrightarrow2\sqrt{6}S\ge\dfrac{4}{a}+a+8a+\dfrac{4}{b}+b+8b+\dfrac{4}{c}+c+8c+ab+bc+ca\)
Áp dụng bất đẳng thức Cauchy
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{4}{a}+a\ge2\sqrt{4}=4\\\dfrac{4}{b}+b\ge2\sqrt{4}=4\\\dfrac{4}{c}+c\ge2\sqrt{4}=4\end{matrix}\right.\)
\(\Rightarrow\dfrac{4}{a}+a+8a+\dfrac{4}{b}+b+8b+\dfrac{4}{c}+c+8c+ab+bc+ca\ge12+8a+8b+8c+ab+bc+ac\)
\(\Rightarrow2\sqrt{6}S\ge12+8a+8b+8c+ab+bc+ac\)
\(\Leftrightarrow2\sqrt{6}S\ge12+2a+bc+2b+ac+2c+ab+6\left(a+b+c\right)\)
Áp dụng bất đẳng thức Cauchy
\(\Rightarrow2a+bc\ge2\sqrt{2abc}\)
Tượng tự ta có \(2b+ac\ge2\sqrt{2abc}\) ; \(2c+ab\ge2\sqrt{2abc}\)
\(\Rightarrow12+2a+bc+2b+ac+2c+ab+6\left(a+b+c\right)\ge6\left(a+b+c+\sqrt{2abc}\right)+12\)
\(\Rightarrow2\sqrt{6}S\ge6\left(a+b+c+\sqrt{2abc}\right)+12\)
Theo đề bài ta có \(a+b+c+\sqrt{2abc}\ge10\)
\(\Rightarrow6\left(a+b+c+\sqrt{2abc}\right)+12\ge72\)
\(\Rightarrow S\ge\dfrac{72}{2\sqrt{6}}=6\sqrt{6}\) ( đpcm )
Dấu " = " xảy ra khi \(a=b=c=2\)
\(A=\dfrac{a}{b+c}+\dfrac{b+c}{a}+\dfrac{b}{c+a}+\dfrac{c+a}{b}+\dfrac{c}{a+b}+\dfrac{a+b}{c}\)
\(A=\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)+\left(\dfrac{b+c}{a}+\dfrac{c+a}{b}+\dfrac{a+b}{c}\right)\)
\(A\ge\dfrac{3}{2}+\dfrac{b+c}{a}+\dfrac{c+a}{b}+\dfrac{a+b}{c}\) (bất đẳng thức Nesbit)
\(A\ge\dfrac{3}{2}+\dfrac{b}{a}+\dfrac{c}{a}+\dfrac{c}{b}+\dfrac{a}{b}+\dfrac{a}{c}+\dfrac{b}{c}\)
\(A\ge\dfrac{3}{2}+\left(\dfrac{b}{a}+\dfrac{a}{b}\right)+\left(\dfrac{c}{a}+\dfrac{a}{c}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)\)
Áp dụng bất đẳng thức AM-GM cho 2 số dương ta có:
\(A\ge\dfrac{3}{2}+2\sqrt{\dfrac{ab}{ab}}+2\sqrt{\dfrac{ac}{ac}}+2\sqrt{\dfrac{bc}{bc}}\)
\(A\ge\dfrac{3}{2}+2+2+2=\dfrac{15}{2}\left(đpcm\right)\)
Dấu"=" xảy ra khi: \(a=b=c\)
Ta chứng minh được:
\(\left(\dfrac{ab}{c}+\dfrac{bc}{a}+\dfrac{ca}{b}\right)^2\ge3\left(a^2+b^2+c^2\right)\)
Thật vậy, bđt đúng với \(\left(\dfrac{ab}{c};\dfrac{bc}{a};\dfrac{ca}{b}\right)=\left(x;y;z\right)\)
\(\left(x+y+z\right)^2\ge3\left(xy+yz+xz\right)\)
\(\Leftrightarrow x^2+y^2+z^2\ge xy+yz+xz\)
\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\)
Đẳng thức xảy ra khi x=y=z=> BĐT cần chứng minh xảy ra dấu bằng khi a=b=c
\(\Rightarrow\dfrac{ab}{c}+\dfrac{bc}{a}+\dfrac{ca}{b}\ge3\)
ta có \(a^2+b^2+c^2\ge3\sqrt[3]{a^2b^2c^2}\Leftrightarrow1\ge\sqrt[3]{a^2b^2c^2}\)
a) theo bđt cauchy schwarz ta có
\(\dfrac{a^3b^3}{c}+\dfrac{b^3c^3}{a}+\dfrac{c^3a^3}{b}\ge3\sqrt[3]{\dfrac{a^6b^6c^6}{abc}}=3\dfrac{a^2b^2c^2}{\sqrt[3]{abc}.1}\ge3\dfrac{a^2b^2c^2}{\sqrt[3]{a^3b^3c^3}}=3abc\)
a) Sai với \(a=1,b=2\)
b)
Thực hiện biến đổi tương đương:
\(\frac{a}{3b}+\frac{b(a+b)}{a^2+ab+b^2}\geq 1\)
\(\Leftrightarrow \frac{a}{3b}+\frac{b(a+b)+a^2}{a^2+ab+b^2}-\frac{a^2}{a^2+ab+b^2}\geq 1\)
\(\Leftrightarrow \frac{a}{3b}-\frac{a^2}{a^2+ab+b^2}\geq 0\)
\(\Leftrightarrow \frac{1}{3b}-\frac{a}{a^2+ab+b^2}\geq 0\)
\(\Leftrightarrow \frac{a^2+ab+b^2-3ab}{3b(a^2+ab+b^2)}\geq 0\)
\(\Leftrightarrow \frac{(a-b)^2}{3b(a^2+ab+b^2)}\geq 0\) (luôn đúng)
Do đó ta có đpcm. Dấu bằng xảy ra khi $a=b$
c) BĐT sai với \(a=1,b=2\)
Áp dụng BĐT AM-GM ta có:
\(\dfrac{1}{a+1}\ge1-\dfrac{1}{b+1}+1-\dfrac{1}{c+1}+1-\dfrac{1}{d+1}\)
\(=\dfrac{b}{b+1}+\dfrac{c}{c+1}+\dfrac{d}{d+1}\)\(\ge3\sqrt[3]{\dfrac{bcd}{\left(b+1\right)\left(c+1\right)\left(d+1\right)}}\)
Tương tự cho 3 BĐT còn lại cũng có:
\(\dfrac{1}{1+b}\ge3\sqrt[3]{\dfrac{acd}{\left(a+1\right)\left(c+1\right)\left(d+1\right)}};\dfrac{1}{c+1}\ge3\sqrt[3]{\dfrac{abd}{\left(a+1\right)\left(b+1\right)\left(d+1\right)}};\dfrac{1}{d+1}\ge3\sqrt[3]{\dfrac{abc}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}}\)
Nhân theo vế 4 BĐT trên ta có:
\(\dfrac{1}{\left(a+1\right)\left(b+1\right)\left(c+1\right)\left(d+1\right)}\ge81\sqrt[3]{\left(\dfrac{abcd}{\left(a+1\right)\left(b+1\right)\left(c+1\right)\left(d+1\right)}\right)^3}\)
\(\Leftrightarrow1\ge81abcd\Leftrightarrow abcd\le\dfrac{1}{81}\)
Ta có :
\(\sqrt{\dfrac{c}{a+b}}+\sqrt{\dfrac{b}{c+a}}+\sqrt{\dfrac{a}{b+c}}=\dfrac{c}{\sqrt{c\left(a+b\right)}}+\dfrac{b}{\sqrt{b\left(c+a\right)}}+\dfrac{a}{\sqrt{a\left(b+c\right)}}\)Áp dụng BĐT Cauchy :
\(\Rightarrow\dfrac{c}{\sqrt{c\left(a+b\right)}}+\dfrac{b}{\sqrt{b\left(c+a\right)}}+\dfrac{a}{\sqrt{a\left(b+c\right)}}\ge\dfrac{2c}{a+b+c}+\dfrac{2b}{a+b+c}+\dfrac{2a}{a+b+c}=2\)Đấu đẳng thức xảy ra khi \(\left\{{}\begin{matrix}a=b+c\\b=c+a\\c=a+b\end{matrix}\right.\)\(\Rightarrow a+b+c=2\left(a+b+c\right)\Rightarrow1=2\) Vậy dấu đẳng thức không xảy ra
Ta phải chứng minh :
\(\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}< 2\)
ta chứng minh bất đẳng thức phụ sau :
do \(\dfrac{a}{a+b}< 1\Rightarrow\dfrac{a}{a+b}< \dfrac{a+c}{a+b+c}\)
tương tự : \(\dfrac{b}{b+c}< \dfrac{b+a}{a+b+c}\); \(\dfrac{c}{c+a}< \dfrac{c+b}{a+b+c}\)
cộng ba vế BĐT lại ta có đpcm
Ta có a,b,c > 0
Áp dụng bất đẳng thức Cô-si : \(a+b\ge2\sqrt{ab}\)
và \(a+b+c\ge3\sqrt[3]{abc}\)
Ta được: Vế trái \(\ge\dfrac{2\sqrt{ab}}{c}+\dfrac{2\sqrt{bc}}{a}+2\dfrac{\sqrt{ac}}{b}\)
\(\ge3\sqrt[3]{\dfrac{2\sqrt{ab}\times2\sqrt{bc}\times2\sqrt{ac}}{abc}}\)
\(\ge3\sqrt[3]{\dfrac{8\sqrt{a^2b^2c^2}}{abc}\ge6}\) (Đpcm)
Vậy: \(\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{c+a}{b}\ge6\)
Đặt A=\(\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{a+c}{b}\)
\(=\left(\dfrac{a}{c}+\dfrac{c}{a}\right)+\left(\dfrac{b}{a}+\dfrac{a}{b}\right)+\left(\dfrac{c}{b}+\dfrac{b}{c}\right)\)
Do a,b,c dương.Áp dụng bất đăng thức côsi cho 2 số dương ta có:
\(\dfrac{a}{b}+\dfrac{b}{a}\ge2\sqrt{\dfrac{ab}{ba}}=2\)
\(\dfrac{b}{c}+\dfrac{c}{b}\ge2\sqrt{\dfrac{bc}{cb}}=2\)
\(\dfrac{c}{a}+\dfrac{a}{c}\ge2\sqrt{\dfrac{ca}{ac}}=2\)
Dấu "=" xảy ra khi và chỉ khi:
\(\dfrac{a}{b}=\dfrac{b}{a};\dfrac{b}{c}=\dfrac{c}{b};\dfrac{c}{a}=\dfrac{a}{c}\Leftrightarrow a=b=c\)
=> A\(\ge6\) ,dấu "=" xảy ra tại a=b=c(đpcm)