MathLabs

Problem 4

Find all positive real solutions (x1,x2,x3,x4,x5)(x_1,x_2,x_3,x_4,x_5) of (x12−x3x5)(x22−x3x5)≤0(x_1^2-x_3x_5)(x_2^2-x_3x_5)\le0, (x22−x4x1)(x32−x4x1)≤0(x_2^2-x_4x_1)(x_3^2-x_4x_1)\le0, (x32−x5x2)(x42−x5x2)≤0(x_3^2-x_5x_2)(x_4^2-x_5x_2)\le0, (x42−x1x3)(x52−x1x3)≤0(x_4^2-x_1x_3)(x_5^2-x_1x_3)\le0, and (x52−x2x4)(x12−x2x4)≤0(x_5^2-x_2x_4)(x_1^2-x_2x_4)\le0.
Step 2 of 4: Regroup as squares
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In plain words

A nonnegative sum cannot be negative; the inequality forces it to be zero.

2∑Ii=∑cyc(xixi+1−xixi+3)2+∑cyc(xixi+2−xixi+4)2≤02\sum I_i=\sum_{cyc}(x_ix_{i+1}-x_ix_{i+3})^2+\sum_{cyc}(x_ix_{i+2}-x_ix_{i+4})^2\le0
Detailed analysis

Expanding, multiplying by 22, and regrouping gives the displayed sum of ten squares. Every term on the right is nonnegative, while the whole sum is at most zero.