Minimum of $(a+\frac{1}{a})^{50}+(b+\frac{1}{b})^{50}+(c+\frac{1}{c})^{50}$ where $a+b+c=1$

Question

Find the minimum of $(a+\frac{1}{a})^{50}+(b+\frac{1}{b})^{50}+(c+\frac{1}{c})^{50}$ where $a+b+c=1$

I tried with the Lagrange Multiplier :

$$\mathcal{L}(a,b,c,\lambda)=(a+\frac{1}{a})^{50}+(b+\frac{1}{b})^{50}+(c+\frac{1}{c})^{50}+\lambda(a+b+c-1)$$

Then

$$\begin{align}\frac{\partial\mathcal{L}}{\partial a}&=50(1-\frac{1}{a^2})(a+\frac{1}{a})^{49}+\lambda=0\\\frac{\partial\mathcal{L}}{\partial b}&=50(1-\frac{1}{b^2})(b+\frac{1}{b})^{49}+\lambda=0\\\frac{\partial\mathcal{L}}{\partial c}&=50(1-\frac{1}{c^2})(c+\frac{1}{c})^{49}+\lambda=0\end{align}$$

So

$$\frac{\partial\mathcal{L}}{\partial a}=\frac{\partial\mathcal{L}}{\partial b}=\frac{\partial\mathcal{L}}{\partial c}$$

But I'm not sure how to make progress after this point. Maybe I can look at the original function and say $(a+\frac{1}{a})^{50}=(a+\frac{1}{a})(a+\frac{1}{a})^{49}$ and substitute an expression in terms of $b$ and $c$, but it doesn't look like this arrives at the minimum.

Answer 1

As

$$ L_{x_k}=n\left(1-\frac{1}{x_k^2}\right)\left(x_k+\frac{1}{x_k}\right)^{n-1}+\lambda = 0 $$

we have as well

$$ nx_k\left(1-\frac{1}{x_k^2}\right)\left(x_k+\frac{1}{x_k}\right)^{n-1}+\lambda x_k = 0 $$

then adding up to $k=1,2,3$ we have

$$ \lambda = -n \sum_k x_k\left(1-\frac{1}{x_k^2}\right)\left(x_k+\frac{1}{x_k}\right)^{n-1} $$

so $$ n\left(1-\frac{1}{x_k^2}\right)\left(x_k+\frac{1}{x_k}\right)^{n-1}-n \sum_i x_i\left(1-\frac{1}{x_i^2}\right)\left(x_i+\frac{1}{x_i}\right)^{n-1}=0 $$

for $i=1,2,3$. Now calling $f(x) = \left(1-\frac{1}{x^2}\right)\left(x+\frac{1}{x}\right)^{n-1}$ we have

$$ \cases{ (1-x_1)f(x_1)-x_2f(x_2)-x_3f(x_3)=0\\ -x_1f(x_1)+(1-x_2)f(x_2)-x_3f(x_3)=0\\ -x_1f(x_1)-x_2f(x_2)+(1-x_3)f(x_3)=0\\ } $$

or

$$ \left(\matrix{1-x_1&-x_2&-x_3\\ -x_1&1-x_2&-x_3\\ -x_1&-x_2&1-x_3}\right)\left(\matrix{f(x_1)\\ f(x_2)\\ f(x_3)}\right)=0 $$

but

$$ \det\left(\matrix{1-x_1&-x_2&-x_3\\ -x_1&1-x_2&-x_3\\ -x_1&-x_2&1-x_3}\right)=1-x_1-x_2-x_3=0 $$

so for continuity $f(x_1)=f(x_2)=f(x_3) = 0$ and the possible answers are

$$ (-1,1,1),(1,-1,1),(1,1,-1) $$

Answer 2

The function $x \mapsto x + \frac1x$ on $x>0$ has a global minimum of $2$ at $x=1$.

The function $y \mapsto y^{50}$ is strictly increasing on $y \geq 2$, so it follows that $x\mapsto \left(x+\frac1x\right)^{50}$ on $x>0$ has a global minimum of $2^{50}$ at $x=1$.

Finally, $x\mapsto \left(x+\frac1x\right)^{50}$ is even, so it has a global minimum of $2^{50}$ at $x=\pm 1$.

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Turning to your problem, this means $$\left(a+\frac1a\right)^{50} + \left(b+\frac1b\right)^{50} + \left(c+\frac1c\right)^{50}$$ has global minima of $3\cdot 2^{50}$ at $a,b,c \in \{-1, 1\}$.

As luck would have it, $-1 + 1 + 1 = 1$, so this unrestricted global minimum is also the constrained minimum.