Consider the system of equations $$x_{1}+x^2_{2}+x^3_{3}+x^4_{4}+x^5_{5} =5$$
and $$x_{1}+2x_{2}+3x_{3}+4x_{4}+5x_{5}=15$$
where $x_{1},x_{2},x_{3},x_{4},x_{5}$ are positive real numbers, Then number of $(x_{1},x_{2},x_{3},x_{4},x_{5})$ is
My Try: From inspection $x_{1}=x_{2}=x_{3}=x_{4}=x_{5}=1$
could some help me how to prove that other solution does not exists, thanks
On
We will use the strict version of Bernoulli's inequality.
For every integer $r \ge 0$ and real number $\epsilon \ge -1$, we have $$(1+\epsilon)^r \ge 1 + r\epsilon$$ and the inequality is strict when $r \ge 2$ and $\epsilon \ne 0$.
For $i = 1,\ldots, 5$, rewrite $x_k$ as $1 + \epsilon_k$. In terms of $\epsilon_k$, the second conditions becomes
$$\sum_{k=1}^5 kx_k = 15\iff \sum_{k=1}^5 k(1+\epsilon_k) = 15 \iff \sum_{k=1}^5 k\epsilon_k = 0\tag{*1}$$
For the first condition, we have
$$\begin{align}\sum_{k=1}^5 x_k^k = 5 & \iff \sum_{k=1}^5 ((1+\epsilon_k)^k - 1 ) = 0\\ & \implies \sum_{k=1}^5 ((1+\epsilon_k)^k - (1+k\epsilon_k)) = 0\\ &\iff \sum_{k=2}^5 ((1+\epsilon_k)^k - (1+k\epsilon_k)) = 0 \end{align} $$
Notice all $x_k > 0$, we have all $\epsilon_k > -1$. We can apply the strict version of Bernoulli's inequality and conclude $\epsilon_2 = \epsilon_3 = \epsilon_4 = \epsilon_5 = 0$. Together with $(*1)$, we find $\epsilon_1 = 0$ too. As a result, the only possible solution has all $\epsilon_k = 0$ which is equivalent to all $x_k = 1$.
On
Consider the functions $f,g:\Bbb R^4\to \Bbb R$ defined by $$x_1=f(x_2,x_3,x_4,x_5)=5-x_2^2-x_3^3-x_4^4-x_5^5$$and $$x_1= g(x_2,x_3,x_4,x_5)=15-2x_2-3x_3-4x_4+5x_5$$
As you have noted, at the point $y=(1,1,1,1)$, $$x_1=f(y)=g(y)=1$$ But note also that $$\nabla f(y)=\nabla g(y)=(-2,-3,-4,-5)$$ that is, they share a tangent plane at $y$. Since $g$ is already a linear function over $\mathbb{R}^4$ plus a constant, it is the tangent plane to the function $f$ at $y$.
Since $f$ is readily seen to be strictly concave (it’s a sum of strictly concave functions in each dimension) on the domain in question, it must lie below its tangent plane $x_1=g$, meeting only at the point $y$.
By AM-GM $$15=x_{1}+2x_{2}+3x_{3}+4x_{4}+5x_{5}\leq x_1+1+x_2^2+2+x_3^3+3+x_4^4+4+x_5^5,$$ which gives $$x_1+x_2^2+x_3^3+x_4^4+x_5^5\geq5.$$ The equality occurs for $x_1=x_2=x_3=x_4=x_5=1$ only, which gives an unique solution.