<s>[INST] Using Lagrange Multipliers to find the maximum of a asymmetric value This problem is from Korean Mathematical Olympiad 2015 P3.
The problem asks to find, with proof, the maximum value of $$(ax+by)^2+(bx+cy)^2$$ with the constraint of $$a^2+b^2+c^2+x^2+y^2=1$$
Now, I do know a solution with AM-GM and Cauchy-Schwarz - I posted on the link.
However, I want to know how to solve this problem using Lagrange Multipliers.
I managed to set up six equations - since there are $5$ variables $a,b,c,x,y$ and also $\lambda$.
Doing that, I got the following equations.
\begin{align*}a^2+b^2+c^2+x^2+y^2&=1\\
ax^2+bxy+a\lambda&=0\\
cy^2+bxy+c\lambda&=0\\
bx^2+axy+cxy+by^2+b\lambda&=0\\
a^2x+aby+bcy+b^2x+x\lambda&=0\\
c^2y+abx+bcx+b^2y+y\lambda&=0\end{align*}
I tried to prove $a=c$ and $x=y$ from there, but I couldn't do it.
How should I solve this monstrous system of equations? Thanks!
 [/INST] Let's say that you guess $a=c$, e.g. by observing equations (2) and (3) as you have them above. Then from equations (5) and (6) you may see that $x=y$ works. For $a=c$ and $x=y$ equation (2) becomes equal to equation (3) and (5) equal to (6), and so you get the system of equations: 
\begin{align*}2a^2+b^2+2x^2&=1\\
ax^2+bx^2+a\lambda&=0\\
2ax^2+2bx^2+b\lambda&=0\\
a^2x+2abx+b^2x+x\lambda&=0
\end{align*}
which may be rewritten as 
\begin{align*}2a^2+b^2+2x^2&=1\\
(a+b)x^2+a\lambda&=0\\
2(a+b)x^2+b\lambda&=0\\
((a+b)^2+\lambda)x&=0
\end{align*}
Getting an expression for $x^2$ from the first equation and substituting in equations (2) and (3), and deriving that $λ=-(a+b)^2$ (or $x=0$) from equation (4), should give you a solution for $a$ and $b$.
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