=>(ab-1)^2+ab(a-b)^2>=0
=>a^2b^2-2ab+1+ab(a^2-2ab+b^2)>=0
=>a^2b^2-2ab+1+a^3b-2a^2b^2+ab^3>=0
=>a^3b+ab^3-a^2b^2-2ab+1>=0
=>ab(a^2+b^2)-2ab-a^2b^2+1>=0
=>ab(a^2+b^2-2-ab)+1>=0(luôn đúng)
=>(ab-1)^2+ab(a-b)^2>=0
=>a^2b^2-2ab+1+ab(a^2-2ab+b^2)>=0
=>a^2b^2-2ab+1+a^3b-2a^2b^2+ab^3>=0
=>a^3b+ab^3-a^2b^2-2ab+1>=0
=>ab(a^2+b^2)-2ab-a^2b^2+1>=0
=>ab(a^2+b^2-2-ab)+1>=0(luôn đúng)
Cho \(a\in R\) . CMR: \(\dfrac{\left(a-4\right)^2}{a^2+16}+\dfrac{\left(a+2\right)^2}{a^2+8}\ge\dfrac{3}{2}\)
usechatgpt init successCho \(x,y>0;x+y=1\) . Tìm Min \(P=\left(x^2+\dfrac{1}{y^2}\right)\left(y^2+\dfrac{1}{x^2}\right)-\dfrac{17}{6}\)
usechatgpt init success\(1,Cho.a,b,c\ge1.CMR:\left(a-\dfrac{1}{b}\right)\left(b-\dfrac{1}{c}\right)\left(c-\dfrac{1}{a}\right)\ge\left(a-\dfrac{1}{a}\right)\left(b-\dfrac{1}{b}\right)\left(c-\dfrac{1}{c}\right)\)
2, Cho a,b,c>0.CMR:
\(\dfrac{a+b}{bc+a^2}+\dfrac{b+c}{ac+b^2}+\dfrac{c+a}{ab+c^2}\le\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)
CMR:
a,\(\dfrac{a^2+b^2}{2}\ge\dfrac{\left(a+b\right)^2}{2}\)
b,Cho a+b=1,a>0,b>0 CMR:\(\left(1+\dfrac{1}{a}\right)\left(1+\dfrac{1}{b}\right)\)\(\ge9\)
B1: Cho \(0\le a,b,c\le2\) thỏa mãn \(a+b+c=3\). CMR: \(a^2+b^2+c^2\le5\)
B2: Cho \(a,b\ge0\) thỏa mãn \(a^2+b^2=a+b\). TÌm GTLN \(S=\dfrac{a}{a+1}+\dfrac{b}{b+1}\)
B3: CMR: \(\dfrac{1}{\left(x-y\right)^2}+\dfrac{1}{x^2}+\dfrac{1}{y^2}\ge\dfrac{4}{xy}\forall x\ne y,xy\ne0\)
1, x,y,z∈N*. CMR x+3z-y là hợp số biết `x^2+y^2=z^2`
2,Tìm n∈N* để \(\left(4n^3+n+3\right)⋮\left(2n^2+n+1\right)\)
3, CMR:\(\dfrac{1}{\left(x-y\right)^2}+\dfrac{1}{x^2}+\dfrac{1}{y^2}\ge\dfrac{4}{xy}\forall x\ne y,xy\ne0\)
Cho a,b,c dương thoả mãn: abc≥1. CMR:
\(\left(a+\dfrac{1}{a+1}\right).\left(b+\dfrac{1}{b+1}\right).\left(c+\dfrac{1}{c+1}\right)\ge\dfrac{27}{8}\)
Chứng minh rằng: \(\left(a^2+b^2+c^2\right)\left[\left(\dfrac{1}{\left(a-b\right)^2}+\dfrac{1}{\left(b-c\right)^2}+\dfrac{1}{\left(c-a\right)^2}\right)\right]\ge\dfrac{9}{2}\)
1. Chứng minh:
\(4-3x+\dfrac{9}{2-3x}\ge8,\forall x< \dfrac{2}{3}\)
2. Cho: a, b, c, d >0 và \(\dfrac{1}{1+a}+\dfrac{1}{1+b}+\dfrac{1}{1+c}+\dfrac{1}{1+d}\ge3\)
Chứng minh rằng: \(abcd\le\dfrac{1}{81}\)
3. Chứng minh rằng: \(\left(\sqrt{a}+\sqrt{b}\right)^8\ge64ab\left(a+b\right)^2,\forall a,b\ge0\)