Question

cho a,b,c>0 thỏa mãn abc=1

cmr: A=\(\dfrac{1}{\sqrt{1+8a}}\) + \(\dfrac{1}{\sqrt{1+8b}}\) +\(\dfrac{1}{\sqrt{1+8c}}\) \(>=\) 1

Answer 1

Đặt \(a^{\dfrac{1}{9}};b^{\dfrac{1}{9}};c^{\dfrac{1}{9}}\rightarrow x;y;z\)\(\left(x;y;z>0;xyz=1\right)\)

Ta có BĐT:\(\dfrac{1}{\sqrt{8x^9+1}}\ge\dfrac{1}{x^8+x^4+1}\)

\(\Leftrightarrow\dfrac{\dfrac{\left(x-1\right)^2x^4\left(x^{10}+2x^9+3x^8+4x^7+7x^6+10x^5+13x^4+8x^3+6x^2+4x+2\right)}{\left(x^2-x+1\right)^2\left(x^2+x+1\right)^2\left(2x^3+1\right)\left(x^4-x^2+1\right)^2\left(4x^6-2x^3+1\right)}}{\dfrac{1}{\sqrt{8x^9+1}}+\dfrac{1}{x^8+x^4+1}}\ge0\)

Tương tự cho 2 BĐT còn lại rồi cộng theo vế:

\(A\ge\dfrac{1}{x^8+x^4+1}+\dfrac{1}{y^8+y^4+1}+\dfrac{1}{z^8+z^4+1}\ge1\)

Dấu "=" khi \(x=y=z=a=b=c=1\)