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If a,b,c are non-negative real numbers for which holds that $a+b+c=3$ then prove the following inequality: $$\frac{9}4\ + \frac{3abc}4\ \ge ab+bc+ca$$

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    $\begingroup$ What did you try ? What are your thoughts? $\endgroup$ Nov 18, 2014 at 12:33
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    $\begingroup$ Hint: Equality is observed when $a=b=c$ or when $a=b, c=0$, which suggests Schur's inequality may be of use. Try it out... $\endgroup$
    – Macavity
    Nov 18, 2014 at 12:57
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    $\begingroup$ Indeed, after trying Schur's inequality I came up with a soultion. Thanks a lot. $\endgroup$
    – CryoDrakon
    Nov 18, 2014 at 13:13
  • $\begingroup$ Great. You could post it as an answer and accept the same, so that others may also benefit. $\endgroup$
    – Macavity
    Nov 18, 2014 at 13:14

2 Answers 2

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If we consider the homogeneous form of our inequality: $$\frac{9}{4}\left(\frac{a+b+c}{3}\right)^3+\frac{3}{4}abc\geq (ab+ac+bc)\left(\frac{a+b+c}{3}\right)$$ we just have to prove that for any triple $(a,b,c)$ of non-negative real numbers $$ a^3+b^3+c^3+3abc \geq ab(a+b)+bc(b+c)+ac(a+c)\tag{1} $$ holds, but $(1)$ is exactly Schur's inequality.

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    $\begingroup$ Your proof is more "elegant". $\endgroup$
    – CryoDrakon
    Nov 18, 2014 at 13:34
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So, we have Schur's inequality which says that for non-negative real numbers a,b,c the following inequality holds: $a^3+b^3+c^3+3abc\ge ab(a+b)+bc(b+c)+ca(c+a)$. Now using the an identity which says: $a^3+b^3+c^3=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)+3abc$ and the fact that $ab+c+=3$ we have: $3(a^2+b^2+c^2-ab-bc-ca)+3abc+3abc\ge ab(3-c)+bc(3-a)+ca(3-b)$

$3(a^2+b^2+c^2)-3(ab+bc+ca)+6abc\ge 3(ab+bc+ca)-3abc$

$3(a^2+b^2+c^2)+9abc\ge 6(ab+bc+ca)$. Dividing both sides with 3 gives us:

$a^2+b^2+c^2+3abc\ge2(ab+bc+ca)$. Now adding to both sides $2(ab+bc+ca)$:

$(a^2+b^2+c^2+2(ab+bc+ca))+3abc\ge4(ab+bc+ca)$

$(a+b+c)^2+3abc\ge4(ab+bc+ca)$. Using now that $a+b+c=3$ and dividng both sides with 4 we get:

$\frac{9}{4}\ + \frac{3abc}4\ge ab+bc+ca$

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