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Lời giải:
Ta viết lại biểu thức vế trái:
\(\text{VT}=\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}=\left(\frac{a}{c}+\frac{a}{b}\right)+\left(\frac{b}{c}+\frac{b}{a}\right)+\left(\frac{c}{a}+\frac{c}{b}\right)\)
\(=a\left(\frac{1}{b}+\frac{1}{c}\right)+b\left(\frac{1}{c}+\frac{1}{a}\right)+c\left(\frac{1}{a}+\frac{1}{b}\right)\)
Áp dụng BĐT Svac-xơ: \(\frac{1}{b}+\frac{1}{c}\geq \frac{4}{b+c}; \frac{1}{c}+\frac{1}{a}\geq \frac{4}{c+a}; \frac{1}{a}+\frac{1}{b}\geq \frac{4}{a+b}\)
Do đó:
\(\text{VT}\geq a.\frac{4}{b+c}+b.\frac{4}{c+a}+c.\frac{4}{a+b}=4\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)\) (đpcm)
Dấu "=" xảy ra khi $a=b=c$
Có: \(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+3\ge4\left(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\right)\)
<=> \(\left(\frac{a}{b}+1\right)+\left(\frac{b}{c}+1\right)+\left(\frac{c}{a}+1\right)\ge\frac{4a}{a+b}+\frac{4b}{b+c}+\frac{4c}{c+a}\)
<=> \(\frac{a+b}{b}+\frac{b+c}{c}+\frac{c+a}{a}\ge\frac{4a}{a+b}+\frac{4b}{b+c}+\frac{4c}{c+a}\)
<=> \(\frac{a+b}{b}+\frac{b+c}{c}+\frac{c+a}{a}+\frac{4b}{a+b}+\frac{4c}{b+c}+\frac{4a}{c+a}\)
\(\ge\frac{4a}{a+b}+\frac{4b}{b+c}+\frac{4c}{c+a}+\frac{4b}{a+b}+\frac{4c}{b+c}+\frac{4a}{c+a}\)
<=> \(\left(\frac{a+b}{b}+\frac{4b}{a+b}\right)+\left(\frac{b+c}{c}+\frac{4c}{b+c}\right)+\left(\frac{c+a}{a}+\frac{4a}{c+a}\right)\)
\(\ge\left(\frac{4a}{a+b}+\frac{4b}{a+b}\right)+\left(\frac{4b}{b+c}+\frac{4c}{b+c}\right)+\left(\frac{4c}{c+a}+\frac{4a}{c+a}\right)\)
<=> \(\left(\frac{a+b}{b}+\frac{4b}{a+b}\right)+\left(\frac{b+c}{c}+\frac{4c}{b+c}\right)+\left(\frac{c+a}{a}+\frac{4a}{c+a}\right)\ge4+4+4\)
<=> \(\left(\frac{a+b}{b}+\frac{4b}{a+b}\right)+\left(\frac{b+c}{c}+\frac{4c}{b+c}\right)+\left(\frac{c+a}{a}+\frac{4a}{c+a}\right)\ge12\)(1)
Áp dụng Cô-si: (1) đúng.
Vậy Bất đẳng thức ban đầu đúng.
"=" <=> a = b = c.
\(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+3\ge4\left(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{a+c}\right)\)
\(\Leftrightarrow\left(\frac{a}{b}+1\right)+\left(\frac{b}{c}+1\right)+\left(\frac{c}{a}+1\right)\ge4\left(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\right)\)
\(\Leftrightarrow\frac{a+b}{b}-\frac{4a}{a+b}+\frac{b+c}{c}-\frac{4b}{b+c}+\frac{c+a}{a}-\frac{4c}{c+a}\ge0\)
\(\Leftrightarrow\frac{\left(a-b\right)^2}{b\left(a+b\right)}+\frac{\left(b-c\right)^2}{c\left(b+c\right)}+\frac{\left(c-a\right)^2}{a\left(a+c\right)}\ge0\)
Luôn đúng vì a,b,c là các số dương
Dấu "=" xảy ra <=> a=b=c
Bớt 6 ở hai vế BĐT cần chứng minh tương đương:
\(\frac{\left(8c-a-b\right)\left(a-b\right)^2+\left(a+b\right)\left(a+b-2c\right)^2}{4abc}\le\frac{\left(7a+7b-2c\right)\left(a-b\right)^2+\left(a+b+2c\right)\left(a+b-2c\right)^2}{2\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
\(\Leftrightarrow\frac{1}{2}\left(a-b\right)^2\left[\frac{7a+7b-2c}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}-\frac{8c-a-b}{2abc}\right]+\frac{1}{2}\left(a+b-2c\right)^2\left[\frac{a+b+2c}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}-\frac{a+b}{2abc}\right]\ge0\)
Tới phần khó chừa lại cho bạn:V
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\)
=> \(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge\left(a+b+c\right)\left(\frac{9}{a+b+c}\right)=9\left(dpcm\right)\)
\(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=3+\frac{a}{b}+\frac{c}{b}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}.\text{ÁP DỤNG BĐT CÔ SI TA ĐƯỢC:}\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=3+\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)\ge3+2\sqrt{\frac{a}{b}.\frac{b}{a}}+2\sqrt{\frac{bc}{bc}}+2\sqrt{\frac{c}{a}.\frac{a}{c}}=3+2+2+2=9\)
Ta có : \(\left(x+y\right)^2\ge4xy\Rightarrow\frac{x+y}{xy}\ge\frac{4}{x+y}\)
\(\Rightarrow\frac{4}{x+y}\le\frac{1}{x}+\frac{1}{y}\)
\(\Rightarrow\frac{1}{x+y}\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right)\)
Áp dụng ta có :
\(\frac{a}{b+c}=a.\frac{1}{b+c}\le a.\frac{1}{4}\left(\frac{1}{b}+\frac{1}{c}\right)=\frac{1}{4}\left(\frac{a}{b}+\frac{a}{c}\right)\)
Tương tự :
\(\frac{b}{c+a}\le\frac{1}{4}\left(\frac{b}{c}+\frac{b}{a}\right)\)
\(\frac{c}{a+b}\le\frac{1}{4}\left(\frac{c}{a}+\frac{c}{b}\right)\)
\(\Rightarrow\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\le\frac{1}{4}\left(\frac{a}{b}+\frac{a}{c}\right)+\frac{1}{4}\left(\frac{b}{c}+\frac{b}{a}\right)+\frac{1}{4}\left(\frac{c}{a}+\frac{c}{b}\right)\)
\(\Rightarrow\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\le\frac{1}{4}\left(\frac{a}{b}+\frac{a}{c}+\frac{b}{c}+\frac{b}{a}+\frac{c}{a}+\frac{c}{b}\right)\)
\(\Rightarrow4\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)\le\frac{a}{b}+\frac{a}{c}+\frac{b}{c}+\frac{b}{a}+\frac{c}{a}+\frac{c}{b}\)
\(\Rightarrow4\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)\le\frac{a+c}{b}+\frac{a+b}{c}+\frac{b+c}{a}\)
\(\Rightarrow4\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)\le\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}\)
Dấu = xảy ra khi a=b=c
Áp dụng BĐT cô si ta có :
\(\frac{b+c}{a}\ge4.\frac{a}{b+c}\)
\(\frac{c+a}{b}\ge\frac{4b}{c+a}\)
\(\frac{a+b}{c}\ge\frac{ac}{a+b}\)
\(\Rightarrow\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}\ge4.\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)\)
Dấu " = " xảy ra khi a= b = c
\(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\frac{\left(xy+yz+zx\right)^2}{x^2y^2z^2}\)(1) với x+y+z=0. Bạn quy đồng vế trái (1) dc \(\frac{x^2y^2+y^2z^2+z^2x^2}{x^2y^2z^2}=\frac{\left(xy+yz+zx\right)^2-2\left(x+y+z\right)xyz}{x^2y^2z^2}\)