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:
$2\cdot k\cdot a +2\cdot c \cdot b$ = $(k\cdot a + k\cdot a)$ + $(c\cdot b + c\cdot b)=(k\cdot a + c\cdot b)$ + $(k\cdot a + c\cdot b)\ge(k\cdot a + c\cdot b)$ + $(k\cdot b + c\cdot b)$

$$
\begin{aligned}
 2(k a+c b)-(k+c)(a+b) 
= & 2 k a+2 c b-k a-c a-k b-b c \\
= & -k a+c b-c a-k b \\
= & (k-c) a+(c-k) b \\
= & (k-c)(a-b)\\ \geqslant & 0 \\
\end{aligned}
$$
Therefore $$k  a + cb \ge \frac{1}{2} (k+c)(a+b)$$
