Show that $(a^2+3)(b^2+3)(c^2+3)(d^2+3)\geq 256$.

Question

Let $a, b, c, d\geq 0$ s.t. $a+b+c+d=4$.

Show that $(a^2+3)(b^2+3)(c^2+3)(d^2+3)\geq 256$.

I don't know how can I deconditioned the inequality.

Answer 1

This is a constrained optimization problem. The Lagrangian function ${\cal L}(a,b,c,d,\lambda) = (a^2+3)(b^2+3)(c^2+3)(d^2+3)-\lambda(a+b+c+d-4)$ has a single real stationary point, which is attained with $a=b=c=d=1$. Convexity arguments may (or may not... did not check) show that this stationary point is a global minimum, which would prove that when $a+b+c+d=4$ we have $(a^2+3)(b^2+3)(c^2+3)(d^2+3) \ge (1^2+3)(1^2+3)(1^2+3)(1^2+3) = 256$.

Note: The stationary points of ${\cal L}$ are the solutions of the system

$$ \nabla {\cal L} = 0 \Leftrightarrow \left\{\begin{array}{l}a(b^2+3)=b(a^2+3)\\b(c^2+3)=c(b^2+3)\\c(d^2+3)=d(c^2+3)\\a+b+c+d=4\end{array}\right.. $$