Can I not use QM-AM inequality to solve this?Bb w XVv x Yo P H Td d Eip t U DWz uCcu R
I was doing my inequality homework and I encountered the following problem:
Show that $\\forall a,b,c\\in\\mathbb{R}^+$, $$\\sqrt{a^2+b^2}+\\sqrt{b^2+c^2}+\\sqrt{c^2+a^2}\\ge(a+b+c)\\sqrt2.$$
I came up a prove using QM-AM inequality, as follows (you can also try to prove with QM-AM inequality for exercise).
But my problem is, can we not use the QM-AM inequality?
-
$\\begingroup$ Do you mean that you want to solve it without AM-QM? $\\endgroup$ – Arthur 12 hours ago
-
$\\begingroup$ @Arthur Yes, the QM-AM proof is just for reference. $\\endgroup$ – Culver Kwan 12 hours ago
-
2$\\begingroup$ The two answers currently given below use essentially the same inequalities, but do not call them AM-QM. Is that good enough? Because "Do not use AM-QM" is a bit vague of a requirement. $\\endgroup$ – Arthur 12 hours ago
-
$\\begingroup$ @Arthur In my worldview, QM/AM does not exist, only AM/GM. $\\endgroup$ – Parcly Taxel 12 hours ago
-
1$\\begingroup$ Consider a square whose one side has length $a+b+c$ and the other $b+c+a$ :). $\\endgroup$ – Michal Adamaszek 12 hours ago
4 Answers
By using Hölder’s Inequality, $$ \\bigl(a^2+b^2\\bigr)^{\\tfrac{1}{2}}\\bigl(1+1\\bigr)^{\\tfrac{1}{2}}\\ge a+b \\\\ \\sqrt{a^2+b^2\\mathstrut} \\ge \\dfrac{a+b}{\\sqrt{2}}$$ Similarly for others, and end it yourself
-
4$\\begingroup$ This is essentially the same proof. $\\endgroup$ – Arnaud D. 12 hours ago
Following the hint given in the comments, consider the following diagram:
The inequality's LHS is the sum of lengths of diagonals that have been drawn in the diagram. However, this being a path from corner to opposite corner, it must be at least as long as a straight line, which has length equal to the RHS. This completes the proof.
Here is a purely visual proof based on Michal Adamaszek's comment:
By CS:
$\\left(\\sum\\limits_{cyc}\\sqrt{a^2+b^2}\\right)^2=2\\sum\\limits_{cyc}a^2+2\\sum\\limits_{cyc}\\sqrt{a^2+b^2}\\sqrt{c^2+a^2}\\geq 2\\sum\\limits_{cyc}a^2+2\\sum\\limits_{cyc}(ac+ab)=2(a+b+c)^2$
