Lời giải:
\(-A=\frac{a^2}{(a-b)(c-a)}+\frac{b^2}{(a-b)(b-c)}+\frac{c^2}{(c-a)(b-c)}\)
\(=\frac{a^2(b-c)+b^2(c-a)+c^2(a-b)}{(a-b)(b-c)(c-a)}=\frac{a^2b+b^2c+c^2a-(ab^2+bc^2+ca^2)}{-[(a^2b+b^2c+c^2a)-(ab^2+bc^2+ca^2)]}=-1\)
$\Rightarrow A=1$
\(1,Cho.a,b,c\ge1.CMR:\left(a-\dfrac{1}{b}\right)\left(b-\dfrac{1}{c}\right)\left(c-\dfrac{1}{a}\right)\ge\left(a-\dfrac{1}{a}\right)\left(b-\dfrac{1}{b}\right)\left(c-\dfrac{1}{c}\right)\)
2, Cho a,b,c>0.CMR:
\(\dfrac{a+b}{bc+a^2}+\dfrac{b+c}{ac+b^2}+\dfrac{c+a}{ab+c^2}\le\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)
Chứng minh rằng: \(\left(a^2+b^2+c^2\right)\left[\left(\dfrac{1}{\left(a-b\right)^2}+\dfrac{1}{\left(b-c\right)^2}+\dfrac{1}{\left(c-a\right)^2}\right)\right]\ge\dfrac{9}{2}\)
Với a, b, c là những số thực dương thỏa mãn \(\left(a+b\right)\left(b+c\right)\)\(\left(c+a\right)\)=1
Chứng minh rằng \(\dfrac{a}{b\left(b+2c\right)^2}\)+\(\dfrac{b}{c\left(c+2a\right)^2}\)+\(\dfrac{c}{a\left(a+2b\right)^2}\)≥\(\dfrac{4}{3}\)
Chứng minh rằng:
\(\left(\dfrac{a+b}{a-b}\right)^2+\left(\dfrac{b+c}{b-c}\right)^2+\left(\dfrac{c+a}{c-a}\right)^2\ge2\)
tính A \(=\dfrac{\left(a^2+b^2+c^2\right)\left(a+b+c\right)^2+\left(ab+bc+ca\right)^2}{\left(a+b+c\right)^2-\left(ab+bc+ca\right)}\)
Sử dụng BĐT Bunhiacopxki cộng mẫu, lm bài toán sau:
Cho a,b,c là các số thực dương. CMR:
\(\dfrac{2\left(b+c-a\right)^2}{2a^2+\left(b+c\right)^2}+\dfrac{2\left(c+a-b\right)^2}{2b^2+\left(c+a\right)^2}+\dfrac{2\left(a+b-c\right)^2}{2c^2+\left(a+b\right)^2}\ge1\)
Biết a,b,c khác 0. CMR: \(\dfrac{a^2}{\left(b-c\right)^2}+\dfrac{b^2}{\left(c-a\right)^2}+\dfrac{c^2}{\left(a-b\right)^2}\ge2\)
\(\dfrac{a}{1+a^2}\)+\(\dfrac{b}{1+b^2}\)+\(\dfrac{c}{1+c^2}\)=\(\dfrac{2}{\sqrt{\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)}}\)
