từ giả thiết, ta có \(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\)
ta có \(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=4\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=4\)
=>\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=2\left(vi:\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\right)\) (ĐPCM)
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Tam giac ABC co chu vi bang 1, ba canh a,b,c thoa man dang thuc
\(\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}=\frac{3}{2}\)
Chung minh tam giac ABC deu
Cho a,b,c thoa man a+b+c=2018 va \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2018}\).
Chung minh co it nhat mot so la 2018
cho \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\) va\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=2.\)
Chung minh rang a+b+c=abc
Cho a,b,c la cac so nguyen duong thoa man: abc=1. CMR
\(\frac{1}{a^3\left(b+c\right)}+\frac{1}{b^3\left(c+a\right)}+\frac{1}{c^3\left(a+b\right)}\ge\frac{3}{2}\)
cho 2 so a va b thoa man a>_1,b>_1 chung minh : \(\frac{1}{1+a^2}+\frac{1}{1+b^2}\ge\frac{2}{1+ab}\)
cho 3 so duong a;b;c thoa man a+b+c=1.tim GTNN cua:
\(p=\left(a+\frac{1}{a}\right)^2+\left(b+\frac{1}{b}\right)^2+\left(c+\frac{1}{c}\right)^2\)
Cho a+b+c=0 va a,b,c≠0. Chứng minh đẳng thức:
\(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}=\sqrt{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}\)
a) Goi a,b,c la do dai ba canh cua mot tam giac thoa man a3 + b3 + c3 = 3abc. Chung minh do la tam giac deu
b) Cho x y z duong va x+y+z = 1. Chung minh \(\frac{1}{x^2+2yz}+\frac{1}{y^2+2xz}+\frac{1}{z^2+2xy}\ge9\)
Cho a,b,c >0
Chứng minh: \(\frac{1}{a^2+b^2+abc}+\frac{1}{b^2+c^2+abc}+\frac{1}{c^2+a^2+abc}\ge\frac{1}{abc}\)