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Let $a$, $b$ and $c$ be real numbers such that $abc\neq0$. Prove that: $$ \frac{(a+b)^2}{c^2}+\frac{(a+c)^2}{b^2}+\frac{(b+c)^2}{a^2}\geq2 $$ I checked for some values and it seems to be true. But no plausible proof is there. I would love a counterexample, or a solution. But please no incomplete hints, I wish to see a complete solution, if it's true. Thanks.

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    $\begingroup$ Is that $a\times b\times c\neq0$ or $a,b,c\neq0$? $\endgroup$
    – Aapeli
    Jun 17, 2014 at 7:05
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    $\begingroup$ Well they are the same right? If one of them is zero then $abc=0$. $\endgroup$
    – shadow10
    Jun 17, 2014 at 7:07
  • $\begingroup$ Easy way out, the equation is symmetric in the variables, and so the minimum is when all are equal. Setting a=b=c, we get the RHS of inequality to be 12. $\endgroup$
    – tpb261
    Jun 17, 2014 at 7:32
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    $\begingroup$ Take $a,-a,-a$ as a counterexample $\endgroup$
    – DGRasines
    Jun 17, 2014 at 9:42
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    $\begingroup$ @shadow10 Hi: when writing titles, please try to write it so that someone looking for an answer to the same question would find your question. "Only a guess but seems to be true" is a good example of how not to title a post :) $\endgroup$
    – rschwieb
    Jun 17, 2014 at 12:40

3 Answers 3

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Assume that $a^2=\min\{a^2,b^2,c^2\}$. By the Cauchy-Schwarz inequality, we have $$\left(\frac {c+a}{b}\right)^2+\left(\frac {a+b}{c}\right)^2\ge \frac{\left((c+a)+(-a-b)\right)^2}{b^2+c^2}=\frac{(b-c)^2}{b^2+c^2}.$$ On the other hand, $$\left(\frac{b+c}{a}\right)^2> \frac{(b+c)^2}{b^2+c^2}.$$ Therefore, $$\left(\frac{b+c}{a}\right)^2+\left(\frac {c+a}{b}\right)^2+\left(\frac {a+b}{c}\right)^2> \frac{(b+c)^2}{b^2+c^2}+\frac{(b-c)^2}{b^2+c^2}=2.$$ The equality does not hold. Is this proof okay? I have another conjecture, which also seems to be true, under previous condition : $$\frac{(a+b)^2}{c^2}+\frac{(a+c)^2}{b^2}+\frac{(b+c)^2}{a^2}\ge 2+ \frac{10(a+b+c)^2}{3(a^2+b^2+c^2)}$$ where equality holds at $a=b=c$ but no formal proof. Thanks.

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    $\begingroup$ I think your proof is correct, and very nice! :D $\endgroup$
    – f10w
    Jun 17, 2014 at 12:36
  • $\begingroup$ Any idea for the second one? Is it true?? $\endgroup$
    – shadow10
    Jun 17, 2014 at 12:43
  • $\begingroup$ Can you please check it again? Because then $a+b=b+c=0$ and $\frac{(a+c)^2}{b^2}=4>2$ ? $\endgroup$
    – shadow10
    Jun 17, 2014 at 12:58
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    $\begingroup$ I know this is Vasce solution:artofproblemsolving.com/Forum/viewtopic.php?f=52&t=593909 $\endgroup$
    – math110
    Jun 17, 2014 at 15:24
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Partial soln. for $a,b,c \gt 0$ the value is same as for $a,b,c, \lt 0$. In both cases we get terms of the form $(\frac{a}{b})^2+(\frac{b}{a})^2+(\frac{a}{c})^2+(\frac{c}{a})^2+(\frac{c}{b})^2+(\frac{b}{c})^2$. This term $\ge 6$, and the remaining term is $2\left(\frac{ac}{b^2}+\frac{ba}{c^2}+\frac{cb}{a^2}\right)$ which is positive. So, when all are of same sign, total is more than 6.

In the cases of two negative and one positive number, or vice versa, let $a,b \gt 0$ and $c \lt 0$. Then the squared terms still give a minimum of 6. But the cross term becomes:$2\left(\frac{-ac}{b^2}+\frac{ba}{c^2}+\frac{-cb}{a^2}\right) = 2\left(\frac{a^3b^3+(-b^3c^3)+(-c^3a^3)}{(abc)^2}\right)$. The numnerator and denominator AM and GM of $a^3b^3$, $-c^3a^3$ and $-b^3c^3$. But it doesn't seem to be helping.

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Let $a+b+c=3u$, $ab+ac+bc=3v^2$, where $v^2$ can be negative, and $abc=w^3$.

Hence, we need to prove that $$\sum_{cyc}a^2b^2(a+b)^2\geq2a^2b^2c^2$$ or $$\sum_{cyc}(a^4b^2+a^4c^2+2a^3b^3)\geq2a^2b^2c^2.$$ We see that $$\sum_{cyc}(a^4b^2+a^4c^2+2a^3b^3)\geq2a^2b^2c^2+\frac{1}{81}\left(\sum\limits_{cyc}(a^2b+a^2c-2abc)\right)^2$$ is a linear inequality of $w^3$, which says that it's enough to prove the last inequality

for an extremal value of $w^3$, which happens for equality case of two variables.

Since the last inequality is homogeneous and even degree, it's enough assume $b=c=1$,

which gives $$79a^4+170a^3-6a^2+7a+322\geq0,$$ which is obviously true.

Done!

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