\(a.\) Ta có : \(\left(a-b\right)^2\) ≥ \(0\) ∀\(ab\)
⇔ \(a^2+b^2\text{ ≥}2ab\)
\(\text{⇔}a^4+2a^2b^2+b^4\text{≥}4a^2b^2\)
\(\text{⇔}a^4+b^4\text{≥}2a^2b^2\)
\(\text{⇔}a^4+b^4\text{≥ }\dfrac{1}{2}\left(a^2+b^2\right)^2\)
Cmtt , \(a^2+b^2\text{≥ }\dfrac{1}{2}\left(a+b\right)^2 \)
⇒ \(a^4+b^4\text{≥ }\dfrac{1}{8}\left(a+b\right)^4\)
CM : \(\left(a^{10}+b^{10}\right)\left(a^2+b^2\right)\ge\left(a^8+b^8\right)\left(a^4+b^4\right)\)
CMR: \(\left(a^2+b^2\right)\left(a^4+b^4\right)\) ≥ \(\left(a^3+b^3\right)^2\)
CMR: \(4\left(a^3+b^3\right)\) ≥ \(\left(a+b\right)^3\) với a, b > 0
CMR Nếu \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\) thì (a+b)\(\left(a^2+b^2\right)\left(a^4+b^4\right)\left(a^8+b^8\right)\left(a^{16}+b^{16}\right)\left(a^{32}+b^{32}\right)\)= \(a^{64}-b^{64}\)
CMR: \(3\left(a^8+b^8+c^8\right)\ge\left(a^3+b^3+c^3\right)\left(a^5+b^5+c^5\right)\)
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
Chứng minh:a)\(x^2+5x-3\ge\dfrac{-37}{4}\)
b)\(a^2+b^2+c^2\ge ab+bc+ac\)
c)\(8\left(x+\dfrac{1}{2}\right)^2+4\left(x^2+\dfrac{1}{x^2}\right)^2-4\left(x^2+\dfrac{1}{x^2}\right)\left(x+\dfrac{1}{x}\right)^2=\left(x+4\right)^2\)
CMR: \(\left(a+b\right)\left(a^3+b^3\right)\le2\left(a^4+b^4\right)\)
CMR: \(\left(a+b\right)\left(a^3+b^3\right)\) ≤ \(2\left(a^4+b^4\right)\)
