Ta có:(a10+b10)(a2+b2)-(a8+b8)(a4+b4)
=a12+b12+a2b10+a10b2-a12-b12-a8b4-a4b8
=a2b2(a8+b8-a6b2-a2b6)
=a2b2[a6(a2-b2)-b6(a2-b2)]
=a2b2(a2-b2)(a6-b6)
=a2b2(a2-b2)(a2-b2)(a4+a2b2+b4)
=a2b2(a2-b2)2(a4+a2b2+b4)
Do a2b2\(\ge\)0 với mọi a;b
(a2-b2)2\(\ge\)0 với mọi a;b
a4+a2b2+b4>0 với mọi a;b(bình phương thiếu)
=>a2b2(a2-b2)2(a4+a2b2+b4)\(\ge\)0 với mọi a;b
=>(a10+b10)(a2+b2)\(\ge\)(a8+b8)(a4+b4)
Ta có bất đẳng thức Bunhiacopski : \(\left(a^2+b^2\right)\left(x^2+y^2\right)\ge\left(ax+by\right)^2\)
Dấu = xảy ra khi \(\dfrac{a}{x}=\dfrac{b}{y}\)
\(\left[\left(a^5\right)^2+\left(b^5\right)^2\right]\left(a^2+b^2\right)\ge\left(a^6+b^6\right)^2\) (1)
\(\left[\left(a^4\right)^2+\left(b^4\right)^2\right]\left[\left(a^2\right)^2+\left(b^2\right)^2\right]\ge\left(a^6+b^6\right)^2\) (2)
Trừ từng vế của 2 bất đẳng thức (1)(2) ta dược : \(\left[\left(a^5\right)^2+\left(b^5\right)^2\right]\left(a^2+b^2\right)-\left[\left(a^4\right)^2+\left(b^4\right)^2\right]\left[\left(a^2\right)^2+\left(b^2\right)^2\right]\ge\left(a^6+b^6\right)^2-\left(a^6+b^6\right)^2\)
\(\Leftrightarrow\) \(\left(a^{10}+b^{10}\right)\left(a^2+b^2\right)-\left(a^8+b^8\right)\left(a^4+b^4\right)\) \(\ge\) 0
\(\Leftrightarrow\) \(\left(a^{10}+b^{10}\right)\left(a^2+b^2\right)\ge\left(a^8+b^8\right)\left(a^4+b^4\right)\)
Dấu bằng xảy ra khi a=b
