\(8\left(a^4+b^4\right)\ge\left(a+b\right)^4\)
\(\Rightarrow8a^4+8b^4\ge\left(a+b\right)^4\)
\(\Rightarrow8\left(a^2\right)^2+8\left(b^2\right)^2\ge\left(a+b\right)^4\)
\(\Rightarrow\left(a+b\right)^4=b^4+4ab^3+6a^2b^2+4a^3+b+a^4\)
\(\Rightarrow8\left(a^4+b^4\right)\ge\left(a+b\right)^4\)(đpcm)
P/s: dấu "=" chỉ xảy ra khi a = b = 1.
\(Cho:a,b,c\ge0.CMR:3\left(a^4+b^4+c^4\right)\ge\left(a+b+c\right)\left(a^3+b^3+c^3\right)\)
Cho a,b,c >0. CMR \(3\left(a^4+b^4+c^4\right)\ge\left(a+b+c\right)\left(a^3+b^3+c^3\right)\)
Cho \(a,b,c>0.\)\(Cmr:\frac{a^4}{\left(a+b\right)\left(a^2+b^2\right)}+\frac{b^4}{\left(b+c\right)\left(b^2+c^2\right)}+\frac{c^4}{\left(c+a\right)\left(c^2+a^2\right)}\ge\frac{a+b+c}{4}\)
Chứng minh bất đẳng thức
a)\(8\left(a^4+b^4\right)\ge\left(a+b\right)^4\)
b)\(\left(a^2+b^2\right)^2\ge ab\left(a+b\right)^2\)
1, Cho x.y=1; x > y. Chứng minh rằng:
\(\frac{x^2+y^2}{x-y}\ge2\sqrt{2}\)
2, CMR : \(\left(a^{10}+b^{10}\right).\left(a^2+b^2\right)\ge\left(a^8+b^8\right).\left(a^4+b^4\right)\)với mọi a,b
Giúp mình nha
1. CM: \(3\left(a^2+b^2\right)-ab+4\ge2\left(a\sqrt{b^2+1}+b\sqrt{a^2+1}\right)\)
2. CMR: \(a^4+b^4+c^4+1\ge2a\left(ab^2-a+c+1\right)\)
3. Cm: \(\left(a^5+b^5\right)\left(a+b\right)\ge\left(a^4+b^4\right)\left(a+b\right)\)
Cm \(8\left(a^4+b^4\right)\ge\left(a+b\right)^4\)
cm:\(\left(a^{10}+b^{10}\right)\left(a^2+b^2\right)\ge\left(a^8+b^8\right)\left(a^4+b^4\right)\)
ai giải giúp bài này với!
Cho a,b>0.CMR:1)\(a^3+b^3\ge ab\left(a+b\right)\)
2)\(a^4+b^4\ge ab\left(a^2+b^2\right)\)
3)\(a^5+b^5\ge ab\left(a^3+b^3\right)\ge a^2b^2\left(a+b\right)\)
p/s: làm bình thường thôi nhá, đừng áp dụng định lí gì cả
