Let $a$, $b$ and $c$ be real positive numbers. Define a function $f$ on the domain $D = \{(x,~y) \in \Bbb R ^2 | x>0,~y>0 \}$ as follows: $$ f(x,y)=ax^{a+b+c}+by^{a+b+c}+c-(a+b+c)x^ay^b. $$
Prove for $(x, y)\in D$ that $$ ax^{a+b+c}+by^{a+b+c}+c\geq (a+b+c)x^ay^b. $$
By Second partial derivative test I found that $(1,1)$ is the unique critical point and especially the local minimum of $f$. If $(1,1)$ is the global minimum then the inequality is proved. Are we available to prove that?
Added: Any local minimum of a convex function is also a global minimum. I wonder if this fact is useful here?
It's just AM-GM: $$\frac{a}{a+b+c}\cdot x^{a+b+c}+\frac{b}{a+b+c}\cdot y^{a+b+c}+\frac{c}{a+b+c}\cdot1\geq$$
$$\geq\left(x^{a+b+c}\right)^{\frac{a}{a+b+c}}\cdot\left(y^{a+b+c}\right)^{\frac{b}{a+b+c}}\cdot1^{\frac{c}{a+b+c}}=x^ay^b$$