5

Given four real numbers $a, b, c, d$ so that $1\leq a\leq b\leq c\leq d\leq 3$. Prove that $$a^{2}+ b^{2}+ c^{2}+ d^{2}\leq ab+ ac+ ad+ bc+ bd+ cd$$

My solution $$3a- d\geq 0$$ $$\begin{align}\Rightarrow d\left ( a+ b+ c \right )- d^{2}= d\left ( a+ b+ c- d \right ) & = d\left ( 3a- d \right )+ d\left ( \left ( b- a \right )+ \left ( c- a \right ) \right )\\ & \geq b\left ( b- a \right )+ c\left ( c- a \right ) \\ & \geq \left ( b- a \right )^{2}+ \left ( c- a \right )^{2} \\ & \geq \frac{1}{2}\left ( \left ( b- a \right )^{2}+ \left ( c- a \right )^{2}+ \left ( c- a \right )^{2} \right )\\ & \geq \frac{1}{2}\left ( \left ( b- a \right )^{2}+ \left ( c- b \right )^{2}+ \left ( c- a \right )^{2} \right )\\ & = a^{2}+ b^{2}+ c^{2}- ab- bc- ca \end{align}$$ How about you ?

  • Mr. @MichaelRozenberg I'm looking forward to your solution and your QUESTION. –  Aug 24 '20 at 12:35
  • 1
    How do you assume $3a-d \ge 0$? Something seems wrong with the question unless I have understood wrong. Are $a = b = c = 1, d = 4$ valid values or not? – Math Lover Aug 24 '20 at 12:54
  • 1
    For all real values of $a, b$ and $c$, $a^2+b^2+c^2 \ge ab+bc+ca$. Same should hold true for $a, b, c, d$. Are you sure inequality says $\le$ and not $\ge$? – Math Lover Aug 24 '20 at 13:00
  • 2
    @MathLover both case show that inequality is completely wrong. Try $a=b=c=d=1$ –  Aug 24 '20 at 13:03
  • @MathLover while what you said is correct, the inequality is symmetrical, so you can assume either way without loss of generality (e.g, $a\ge d$). – Quang Hoang Aug 24 '20 at 13:34
  • 1
    @QuangHoang of course you can choose any arbitrary values but if you assume $(3a-d) \ge 0$, next set of steps are not correct. Please go through his steps. – Math Lover Aug 24 '20 at 13:52
  • 1
    If $a = 1 + \epsilon$ and $d= 4-\delta$ than $3a - d =-1 + 3\epsilon + 4\delta \ge 0$ only if $3\epsilon + 4 \delta \ge 1$. There is utterly no reason to assume that is true. $a =1$ and $d =4$ is an obvious counter example. But so is $a = 1.2$ and $d= 3.95$. – fleablood Aug 24 '20 at 18:40
  • 1
    "he inequality is symmetrical, so you can assume either way without loss of generality " No. there is a HUGE loss of generality. If you choose an $a$ and $d$ so that $3a-d \ge 0$ you get a very different result than if you choose an $a$ and $d$ where $3a-d < 0$. Yes, your choice are "arbitrary" but they can not be specific. Choosing $a = 2.1; b=2.5; c=3.1; d= 3.7$ is also arbitrary. But they are not general. Choosing them will not prove all cases. There IS a loss of generality. – fleablood Aug 24 '20 at 18:44

1 Answers1

2

It's wrong.

Try $$(a,b,c,d)=(1,1,1,4).$$ For these values we need to prove that $$19\leq15,$$ which is not so true.

The following inequality is true already.

let $\{a,b,c,d\}\subset[1,3].$ Prove that: $$a^2+b^2+c^2+d^2\leq ab+ac+bc+ad+bd+cd.$$

We can prove this inequality by the Convexity.

Indeed, let $f(a)=ab+ac+bc+ad+bd+cd-a^2-b^2-c^2-d^2$.

Thus, $f$ is a concave function, which says that $f$ gets a minimal value for an extreme value of $a$,

id est, for $a\in\{1,3\}$.

Similarly, for $b$, $c$ and $d$.

Thus, it's enough to check our inequality for $\{a,b,c,d\}\subset\{1,3\}$, which gives that our inequality is true.

Michael Rozenberg
  • 203,855