Find the minimizer of $A=12a+13b+11c$ 
Let $a \ge 2$, $b \ge 5$ and $c \ge 5$ such that $2a^2+b^2+c^2=69$. Find the minimizer of $$A=12a+13b+11c$$


My try: Use Lagrange multipliers
We have: $L\left(a;b;c;\lambda \right)=12a+13b+11c+\lambda \left(2a^2+b^2+c^2-69\right)$
$$\begin{cases}  
\frac{\partial L}{\partial a}=4a\lambda +12=0\Leftrightarrow \lambda =-\frac{3}{a} &  \\ 
\frac{\partial L}{\partial b}=2b\lambda +13 =0\Leftrightarrow \lambda =-\frac{13}{2b}    \\
\frac{\partial L}{\partial c}=2c\lambda +11=0\Leftrightarrow \lambda =-\frac{11}{2c} &  
\end{cases} $$ 
Or $$\begin{cases}  
\frac{3}{a}=\frac{13}{2b}=\frac{11}{2c} &  \\ 2a^2+b^2+c^2=69 &  
\end{cases} $$
But $\text{Min}A=155$ occurs when $a=2;b=5;c=6$. Help me!

Let $a=2$, $b=5$ and $c=6$.
Hence, $A=155.$
We'll prove that it's a minimal value.
Indeed, let $a=u+2$, $b=v+5$ and $c=w+5$.
Hence, the condition gives
$$2u^2+v^2+w^2+8u+10v+10w=11$$ and we need to prove that
$$12u+13v+11w\geq11.$$
For the proof we can use the Contradiction method.
Indeed, let $12u+13v+11w<11$, $u=kx$, $v=ky$ and $w=kz$, 
where $k>0$ and $12x+13y+11z=11.$
Thus, $k<1$ and
$$11=2u^2+v^2+w^2+8u+10v+10w=k^2(2x^2+y^2+z^2)+k(8x+10y+10z)<$$
$$<2x^2+y^2+z^2+8x+10y+10z,$$
which is a contradiction because we'll prove now that
$$2x^2+y^2+z^2+8x+10y+10z\leq11.$$
Indeed, we need to prove that
$$2x^2+y^2+z^2+\frac{(8x+10y+10z)(12x+13y+11z)}{11}\leq\frac{(12x+13y+11z)^2}{11}$$ or
$$13x^2+14y^2+44xy+28xz+23yz\geq0,$$ which is obvious.
Done!
