<s>[INST] How to show $k \cdot a + c \cdot b \ge \frac{1}{2} \cdot (k+c) \cdot (a+b)$? Assume k, c, a, b $\gt$ 0 and that $k \ge c$ and $a \ge b$. Prove $k \cdot a$ + $c \cdot b$ $\ge$ $\frac{1}{2}\cdot (k+c) \cdot (a+b)$. This is what I have so far:
$2\cdot k\cdot a +2\cdot c \cdot b$ = $(k\cdot a + k\cdot a)$ + $(c\cdot b + c\cdot b)=(k\cdot a + c\cdot b)$ + $(k\cdot a + c\cdot b)\ge(k\cdot a + c\cdot b)$ + $(k\cdot b + c\cdot b)$
 [/INST] We have $k≥c$ and $a≥b$. And we have to prove that $ka+bc≥(k+c)(a+b)/2$. Now by algebraic expansion we can write it out like these $$ka+bc≥(k+c)(a+b)/2$$ $$or, ka+bc≥kb+ac$$. Since $k≥c$ and $a≥b$ by Appling rearrangement inequality, the result follows.
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