Problem
ALG-B1-M01-P022 Difference of Fourth Powers
#22
★★★★☆ Level 4 of 5
Prove that \(a^4+b^4\ge a^3b+ab^3\) for all real \(a,b\).
Move everything to the left and try to factor out \((a-b)^2\).
Consider the difference: \[ a^4+b^4-a^3b-ab^3=a^3(a-b)-b^3(a-b)=(a-b)(a^3-b^3). \] Also \(a^3-b^3=(a-b)(a^2+ab+b^2)\).
So the difference is \((a-b)^2(a^2+ab+b^2)\). The first factor is non-negative, and the second is \(\left(a+\frac b2\right)^2+\frac34b^2\ge0\). Hence the inequality is proved.
This is olympiad technique: prove an inequality by exact factorization.