I'm not a big fan of inequalities, as I can't really solve one without having a book with theory in front of me. Who knows, maybe I'll make a post about interesting inequalities once.
Even though I'm not really a big fan of inequalities, I am a fan of analysis, and this was my introduction to multivariable calculus, so I enjoyed this problem. Of course, my approach may not be the simplest; when I asked one of my friends over at a beer about this problem, he said that he could have solved it using elementary inequalities. If you can find a solution using only elementary inequalities, I encourage you to write it down below. The problem was taken, yet again, from an AoPS post.
The problem goes something like this:
If 
and 
then prove that 
.
First off, we have to find an interval in which lie all the solutions of the given equations.
Thus, by using CBS,
$$(x+y+z+t)^{2}<4(x^2+y^2+z^2+t^2) $$, which implies that $$(-u)^2<4(20-u^2)$$.
\(u^2<80-4u^2,\) so \(5u^2<80\), and, thus, \(-4<u<4\). Obviously, all the other variables satisfy the same condition. Now we know that
x,y,z,t,u are in [-4,4]. Let us now denote the following functions defined on \([-4,4]^5\) with values in the reals:
$$g_1(x,y,z,t,u) = x+y+z+t+u$$
$$g_2(x,y,z,t,u)=x^2+y^2+z^2+t^2+u^2-20$$
$$f(x,y,z,t,u)=x^4+y^4+z^4+t^4+u^4$$.
We need to find the maximum value of the function $$F=f+ a g_1+ b g_2$$. Since \(f, g_1, g_2\) are derivable with a continuous derivative, the function F has a maximum on the given intervals, and, to find the maximum, we must differentiate F in x, y, z, t and u. In order to find the maximum, we must also use the fact that \(g_1=g_2=0\), which we know from the hypothesis.
After differentiating, we get the following set of equations:
$$4x^3+a+2xb=0$$
$$4y^3+a+2yb=0$$
$$4z^3+a+2zb=0$$
$$4t^3+a+2tb=0$$
$$4u^3+a+2ub=0$$
Since the equation $$4x^3+2xb+c=0$$ has a maximum number of three solutions, we can discern between two cases: One in which there are two pairs of equal numbers among (x,y,z,t,u), and one in which there are 3 equal numbers.
FIRST CASE: x=y, z=t.
We now have that \(2x+2y+u=0\) and \(2x^2+2y^2+u^2=20\). By writing \(u=-2(x+y)\), we get that
\(2x^2+2y^2+4(x+y)^2=20\), and, thus, \(6x^2+8xy+6y^2=20\), so \(3x^2+4xy+3y^2=10\) (1).
By rising to the second power, we get that \((3x^2+4xy+3y^2)^2=100\), and, thus,
$$9x^4+24x^{3}y+34x^{2}y^{2}+24xy^{3}+9y^4=100$$, and, multiplying it by 2, we get that
$$18x^4+48x^{3}y+68x^{2}y^{2}+48xy^{3}+18y^4=200$$
Now, \(x^4+y^4+z^4+t^4+u^4=2x^4+2y^4+16(x^4+4x^{3}y+6x^{2}y^{2}+4xy^{3}+y^4 =\)
\(=18x^4+18y^4+64x^{3}y+64xy^{3}+96x^{2}y^{2}=\)
\(=200+16x^{3}y+16xy^{3}+28x^{2}y^{2}\),
which is smaller or equal to 260 if and only if
$$16x^{3}y+16xy^{3}+28x^{2}y^{2}<60$$and, by dividing by 2,
$$8x^{3}y+8xy^{3}+14x^{2}y^{2}<30$$which can be written as
$$xy(8x^2+8y^2+14xy)<30$$ (*)
We know that \(|u|=|-2(x+y)|<4\), so \(|x+y|<2\). By squaring, we get that \((x+y)^2<4\), and \(x^2+y^2>2xy, so xy<1\) (2).
Since \((x+y)^2=x^2+y^2+2xy<4\),
\(2x^2+4xy+2y^2<8\), while \(3x^2+4xy+3y^2=10\) (from (1)). Thus, we get a new inequality:
\(x^2+y^2>2\) (3)
Now, we write $$14xy=(28xy)/2=[70-21(x^2+y^2)]/2$$. Back to (*), we get that
$$xy(8x^2+8y^2+14xy)<xy[8x^2+8y^2+35-(21/2)(x^2+y^2)]=$$
$$(xy/2)[70+16(x^2+y^2)-21(x^2+y^2)]= (xy/2)[(70-5(x^2+y^2)]$$, which is, according to (3), smaller than $$(xy/2)(70-10)<30xy<30$$, so (*) is proven.
If you have come this far, I suppose you are curious about the second case, too. Writing this is kind of tedious, but fun nonetheless.
SECOND CASE: x=y=z
\(3x+u+t=0\), \(3x^2+u^2+t^2=20\), so \(u+t=-3x\) and \(u^2+t^2+2ut=9x^2\).
We also know that \(u^2+t^2=20-3x^2\), so \(ut=6x^2-10\)
Thus, \(t^4+u^4=(t^2+u^2)^2-2(tu)^2=\)
\(=400-120x^2+9x^4-2(36x^4-120x^2+100)=200+120x^2-63x^4\).
$$x^4+y^4+z^4+t^4+u^4=-60x^4+120x^2+200$$.
We have to find the maximum value of this expression on the interval [-4,4]. Let us write \(A=x^2\). Now, the function we have to evaluate is \(g(A)=-60A^2+120A+200\), which obviously has a maximum value, because it's a grade two function and the coefficient of the greatest power is negative.
By differentiating, we have to find the A for which $$g'(A)=-120A+120=0$$ only when \(A=1\), and, thus, the maximum value of \(x^4+y^4+z^4+t^4+u^4=260\).
Thus, after solving both cases, we can conclude that \(x^4+y^4+z^4+t^4+u^4<260\).
Note that, wherever I wrote <, I meant smaller or equal to something. I am still working on my Latex skills, and, for me, writing the full proof of this inequality was a feat of strength.
I have found another inequality that seems solvable with Lagrange Multipliers a while ago, yet I haven't solved it yet. If I manage to solve it, I'll make sure to post it. If you people have a solution to this problem that doesn't involve Lagrange Multipliers, please write it in the comments. I'm really curious about it :)
Cris.
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