# An Inequality with Constraint VII

### Problem

Dan Sitaru has kindly posted a problem from his book "Liquid Math" at the CutTheKnotMath facebook page:

If $x,y,z\in\mathbb{R},\;$ $x+y-5z=0,\;$ $x^2+z^2=1.\;$ Then

$|2x+3y-5z|\le\sqrt{101}.$

### Solution

Since $x+y-5z=0,\;$ the inequality at hand is equivalent to

$|(2x+3y-5z)-3(x+y-5z)|=|-x+10z|\le\sqrt{101}.$

This is the one I shall prove under the restriction $x^2+z^2=1.\;$ After the fact, $y\;$ may be found from $x+y-5z=0.$

The straight lines $-x+10z=\text{const}\;$ may or may not meet the circle $x^2+z^2=1.\;$

I shall employ geometric illustration. The value of $-x+10z\;$ which is constant on each of the lines changing monotonically in the direction of their common normal: $(-10,1).$ The extreme values are attained at the intersection of $x^2+z^2=1\;$ with $z=-10x:$

This happens when $\displaystyle x=\pm\frac{1}{\sqrt{101}},\;$ $z=\displaystyle\mp\frac{10}{\sqrt{101}}$ such that, at these points, $|-x+10z|=\sqrt{101},\;$ which proves the required inequality.

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