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I have problems with proving inequality :

${a^{2}}+b^2+c^2+\frac{2}{5}abc<50$

where $a,b,c$ are the lengths of triangle's sides, and the circumference of the triangle is $10$.

Thanks.

  • 0
    Presumably by "circumference" what's meant is what I'd call "perimeter".2011-05-15

2 Answers 2

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Consider the polynomial $(x-a)(x-b)(x-c)$. Multiplied out, this is

$ \begin{eqnarray} (x-a)(x-b)(x-c) &=& x^3-(a+b+c)x^2+(ab+ac+bc)x-abc \\ &=& x^3-10x^2+(ab+ac+bc)x-abc\;. \end{eqnarray} $

We also have

$10^2=(a+b+c)^2=(a^2+b^2+c^2) + 2(ab+ac+bc)\;,$ $ab+ac+bc=\frac{100-(a^2+b^2+c^2)}{2}\;,$

and thus

$ (x-a)(x-b)(x-c) = x^3-10x^2+\frac{100-(a^2+b^2+c^2)}{2}x-abc\;. $

Now since $a$, $b$ and $c$ form a triangle with perimeter $10$, they must all be less than $5$. Thus the value of the polynomial for $x=5$ is positive, that is,

$ 5^3-10\cdot5^2+\frac{100-(a^2+b^2+c^2)}{2}\cdot5-abc>0\;, $

which upon rearrangement becomes your inequality.

  • 2
    Bravo!${}{}{}{}$2011-05-15
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We need to prove that $\frac{(a+b+c)(a^2+b^2+c^2)}{10}+\frac{2}{5}abc<\frac{(a+b+c)^3}{20}$ or $(a+b+c)^3>8abc+2(a+b+c)(a^2+b^2+c^2)$ or $\sum_{cyc}(a^3+3a^2b+3a^2c+2abc)>8abc+2\sum_{cyc}(a^3+a^2b+a^2c)$ or $\sum_{cyc}\left(-a^3+a^2b+a^2c-\frac{2}{3}abc\right)>0$ or $(a+b-c)(a+c-b)(b+c-a)>0$ Done!