How can I evaluate the minimum of $$ \left|7x-1\right|+\left|7y-5\right|+\left|7z-1\right| $$ if $x,y,z$ are non negative reals such that $ x+y+z=1$ and $y^2 \le 3xz$?
Without softwares help..
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It seems to me that $x$ and $z$ play the same game. Could $z=x$ ??? – Claude Leibovici Mar 29 '16 at 10:51
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Mmh, most probably you are right; do you think it is possible to formalize it? – Mar 29 '16 at 11:17
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Here is one way to formalise the argument suggested by Claude above. Consider replacing both $x, z$ with $t=\frac{x+z}2$. Clearly the objective and the constraint $x+y+z=1$ are not disturbed, while as $3t^2\ge 3xz$ we now have more slack in the remaining constraint. So to find the optimum, we may set $x=z$ to have the equivalent:
Minimise $2|7x-1|+|7y-5|, \;$ such that $2x+y=1$ and $\sqrt3x \ge y^2.\;$ Eliminate $y$ by setting $y = 1-2x$, so that we have $\sqrt3 x \ge 1-3x \implies x \ge 2-\sqrt3$ and now need to minimise $28|x-\frac17|$.
As $2-\sqrt3 > \frac17$, the minimum will be when $x= z = 2-\sqrt3, \; y = 2\sqrt3-3$ for a minimum value of $52-28\sqrt3 \approx 3.50$.
Macavity
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One way to tackle this problem is to study eight cases, whether each of the expressions under absolute value is positive or not and then to invoke the Lagrange multipliers.
Santiago
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Well, I did know this pretty manual solution; but isn't any better option? – Paolo Leonetti Mar 29 '16 at 10:34