# a,b>1 prove, a^2/(b-1)+b^2/(a-1)≥8?

### 2 Answers

- JOHNLv 72 years agoBest Answer
Clear fractions and substitute a = x + 1, b = y + 1 → x, y > 0;

inequality becomes x(x + 1)² + y(y + 1)² ≥ 8xy; expand LHS:

x³ + y³ + 2x² + 2y² + x + y = x³ + y³ + x² + x² + y² + y² + x + y

≥ 8(x⁸y⁸)^(1/8) = 8(xy) (AM - GM here); given inequality then follows.

- IndikosLv 72 years ago
We repeatedly use the AM_GM inequality

viz. if a and b > 0 then a^2 + b^2 >= 2ab

Also we prove that if x>0 then x + 1/x >=2

Let h=sqrt(x)

By AM-GM inequality we have h^2 + 1/h^2 >= 2h/h = 2 or x + 1/x >= 2

Now a^2/(b-1)+b^2/(a-1)

= ( ( (a-1)^2 + 2(a-1) + 1)/(b-1) * ( (b-1)^2 + 2(b-1) + 1) /(a-1) )

>= 2 sqrt( ( (a-1)^2 + 2(a-1) + 1)/(b-1) * ( (b-1)^2 + 2(b-1) + 1) /(a-1) ) ,,, by AM_GM

= 2 sqrt( ( (a-1)^2 + 2(a-1) + 1)/(a-1) * ( (b-1)^2 + 2(b-1) + 1) /(b-1) )

= 2 sqrt( ( (a-1) + 2 + 1/(a-1) ) * ( (b-1) + 2 + 1 /(b-1) )

>= 2 * sqrt( (2 + 2)(2+2) ) ,,,,,,,,,,, since x+1/x >= 2

= 2 * sqrt(16)

= 8