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Lời giải:
$\frac{1}{c}=-(\frac{1}{a}+\frac{1}{b})< 0$ do $a,b>0$
$\Rightarrow c< 0$
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Leftrightarrow ab+bc+ac=0$
Từ đây ta có:
\((\sqrt{a+c}+\sqrt{b+c})^2=a+c+b+c+2\sqrt{(a+c)(b+c)}\)
\(=a+b+2c+2\sqrt{ab+bc+ac+c^2}=a+b+2c+2\sqrt{c^2}\)
\(=a+b+2c+2|c|=a+b+2c+2(-c)=a+b\)
\(\Rightarrow \sqrt{a+c}+\sqrt{b+c}=\sqrt{a+b}\) (do \(\sqrt{a+c}+\sqrt{b+c}\geq 0\))
Ta có đpcm.
\(\left(1-a\right)\left(1-b\right)\left(1-c\right)\left(1-d\right)\)
\(=abcd+bd+cd+ab\left(1-c\right)+ad\left(1-b\right)+ac\left(1-d\right)+bc\left(1-d\right)+\left(1-a-b-c-d\right)\)
\(>1-a-b-c-d\)
\(N=\frac{a}{1+b^2c}+\frac{b}{1+c^2d}+\frac{c}{1+d^2a}+\frac{d}{1+a^2b}\)
Áp dụng BĐT Cauchy ta có:
\(\frac{a}{1+b^2c}=a-\frac{ab^2c}{1+b^2c}\)
\(\ge a-\frac{ab^2c}{2b\sqrt{c}}=a-\frac{ab\sqrt{c}}{2}=a-\frac{b\sqrt{ac}\sqrt{a}}{2}\)
\(\ge a-\frac{b\left(ac+c\right)}{4}\).Suy ra \(\frac{a}{1+b^2c}\ge a-\frac{1}{4}\cdot\left(ab+abc\right)\)
Tương tự ta có:
\(\frac{b}{a+c^2d}\ge b-\frac{1}{4}\left(bc+bcd\right)\)
\(\frac{c}{1+d^2a}\ge c-\frac{1}{4}\left(cd+cda\right)\)
\(\frac{d}{1+a^2b}\ge d-\frac{1}{4}\left(da+dab\right)\)
Do đó: \(S=\frac{a}{1+b^2c}+\frac{b}{1+c^2d}+\frac{c}{1+d^2a}+\frac{d}{1+a^2b}\)
\(\ge a+b+c+d-\frac{1}{4}\left(ab+bc+cd+da+abc+bcd+cda+dab\right)\)
\(=4-\frac{1}{4}\left(ab+bc+cd+da+abc+bcd+cda+dab\right)\)
Ta có:
\(ab+bc+cd+da\le\frac{1}{4}\left(a+b+c+d\right)^2=4\)
\(abc+bcd+cda+dab\le\frac{1}{16}\left(a+b+c+d\right)^3=4\)
nên \(S\ge4-\frac{1}{4}\cdot\left(4+4\right)=2\)(Đpcm)
Dấu = khi \(a=b=c=d=1\)