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Let a triangle with sides a b and c. $p=\frac{a+b+c}{2}$ Prove that $$\frac{(p-a)(p-b)(p-c)(a^2+b^2+c^2+ab+bc+ca)}{abcp^2}\le\frac{2}{3}$$

My thought: Let $a=x+y,b=y+z,c=z+x$ so $x+y+z=p$ Then let $p=x+y+z,q=xy+yz+zx,r=xyz$ we get $10p^2r\le3qr+p^3q$ I don't know what to do next. Thanks

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    I think your constants are not good. Can you take a look at that?2012-11-07
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    In the title of the thread you have 4 in the denominator, however you don't have 4 in the denominator in the second inequality2012-11-07
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    looks like a homework, buddy2012-11-07

2 Answers 2

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By AM-GM inequality $$ \sqrt{(p - a)(p - b)} = \frac 1 2 \sqrt{(b + c - a)\cdot (a + c - b)}\leq\\ \frac 1 2 \frac {(b + c - a) + (a + c - b)}{2} = \frac 1 2 c $$ Hence $$ (p - a)(p - b)(p - c) = \\ \sqrt{(p - a)(p - b)} \cdot \sqrt{(p - b)(p - c)} \cdot \sqrt{(p - c)(p - a)} \leq \frac 1 8 a b c $$ Using the above inequality we get $$ LHS \leq \frac {a^2 + b^2 + c^2 + ab + bc + cd} {8 p^2} \leq \frac 1 2 + \frac 1 2 \frac {ab + bc + ca} {(a + b +c)^2} \tag 1 $$ Now by Cauchy- Schwarz inequality $$ a\cdot b + b\cdot c + c\cdot a \leq a^2 + b^2 + c^2 $$ we deduce $$ (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca \geq 3 (ab + bc + ca) $$ Substituting the above relation into $(1)$ yields $$ LHS \leq \frac 1 2 + \frac 1 6 = \frac 2 3 $$

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First notice that $p^{2} = (\frac{a+b+c}{2})^{2} = \frac{a^{2}}{4} + \frac{ab}{2} + \frac{ac}{2} + \frac{b^{2}}{4} + \frac{bc}{2} + \frac{c^{2}}{4} < a^2 + b^2 + c^2 + ab + bc + ca$ which should help to simplify the situation. Also $(p-a)(p-b)(p-c) = \frac{b+c}{2} + \frac{a+c}{2} + \frac{a+b}{2} < 2(a+b+c)$ might be of help.