If $a,b,c\in\mathbb{R},$ such that $a\geq 2$, $b\geq 5$, $c\geq 5$ and $2a^2+b^2+c^2=69$. Prove that $12a+13b+11c \geq 155.$ So,I am trying to prove the following inequality without using the Lagrange multipliers:
If $a,b,c\in\mathbb{R},$ such that $a\geq 2$, $b\geq 5$, $c\geq 5$ and $2a^2+b^2+c^2=69$. Prove that
$$12a+13b+11c \geq 155.$$
My try:
$\bullet a\geq 2, b\geq 5, c\geq 5,2a^2+b^2+c^2=69 \rightarrow 2 \le a<4,5 \le b,c \le 6.$
But,I don't know how to proceed further.
Any help would be highly appreciated.Thank you!

Let $a = 2 + x, b = 5 + y, c = 5 + z$. Then $x, y, z \ge 0$.
We have $2a^2 + b^2 + c^2 - 69
= 2x^2 + y^2 + z^2 + 8x + 10y + 10z - 11 = 0$.
We want to prove that $12x + 13y + 11z \ge 11$.
We split into two cases:
(1) If $x > 1$ or $y > 1$ or $z > 1$, we have $12x + 13y + 11z \ge 11$.
(2) If $x, y, z \in [0, 1]$, we have
$$2x^2 + y^2 + z^2 + 8x + 10y + 10z
\le 2x + y + z + 8x + 10y + 10z
= 10x + 11y + 11 z $$
which results in $10x + 11y + 11z \ge 11$.
Thus, we have $12x + 13y + 11z
\ge 10x + 11y + 11 z \ge 11$.
We are done.
