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Given four real numbers $a,b,c,d$ so that $1leq aleq bleq cleq dleq 3$. Prove that $a^2+b^2+c^2+d^2leq ab+ac+ad+bc+bd+cd.$

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

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

How about you ?

Mathematics Asked by user818748 on December 29, 2020

1 Answers

One Answer

It's wrong.

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

The following inequality is true already.

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

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 $ain{1,3}$.

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.

Correct answer by Michael Rozenberg on December 29, 2020

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