Cho a, b, c > 0 . CMR:
\(\frac{1}{a+b+c}\ge\frac{a^3}{\left(2a^2+b^2\right)\left(2a^2+c^2\right)}+\frac{b^3}{\left(2b^2+c^2\right)\left(2b^2+a^2\right)}+\frac{c^3}{\left(2c^2+a^2\right)\left(2c^2+a^2\right)}\)
Cho a + b + c = 0. CMR \(a^4+b^4+c^4=2\left(a^2b^2+b^2c^2+c^2a^2\right)=2\left(ab+bc+ca\right)^2=\dfrac{\left(a^2+b^2+c^2\right)^2}{2}\)
Cho \(a,b,c>0\) và \(a+b+c=3\). CMR : \(P=\dfrac{a^3}{b\left(2c+a\right)}+\dfrac{b^3}{c\left(2a+b\right)}+\dfrac{c^3}{a\left(2b+c\right)}\ge1\)
giúp nha mn :==|_T-T
Cho a+b+c=0.Chứng minh
\(\frac{b-c}{a\left(a-b\right)}\)+\(\frac{c-a}{b\left(a-b\right)}\)=\(\frac{2c}{ab}\)
Cho a,b,c khác 0 thỏa mãn \(\left(1+\dfrac{b}{a}\right)\left(1+\dfrac{c}{b}\right)\left(1+\dfrac{a}{c}\right)=8\)
CMR \(\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}=\dfrac{3}{4}+\dfrac{ab}{\left(a+b\right)\left(b+c\right)}+\dfrac{bc}{\left(b+c\right)\left(c+a\right)}+\dfrac{ca}{\left(c+a\right)\left(a+b\right)}\)
CMR: \(8\left(a^3+b^3+c^3\right)\ge\left(a+b\right)^3+\left(b+c\right)^3+\left(c+a\right)^3\) với a, b, c > 0
Cho \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=4\left(a^2+b^2+c^2-ab-ac-bc\right)\)
So sánh a + b với 2c
Đề thi học sinh giỏi vòng trường lớp 8
Chứng minh với mọi a,b,c > 0, ta có:
\(\dfrac{\left(2b+3c\right)^2}{a}+\dfrac{\left(2c+3a\right)^2}{b}+\dfrac{\left(2a+3b\right)^2}{c}\ge25\left(a+b+c\right)\)
CMR Nếu \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\) thì (a+b)\(\left(a^2+b^2\right)\left(a^4+b^4\right)\left(a^8+b^8\right)\left(a^{16}+b^{16}\right)\left(a^{32}+b^{32}\right)\)= \(a^{64}-b^{64}\)