Any solution?

Can this be useful? $\displaystyle (a+b+c)^5-a^5-b^5-c^5=5(a+b)(b+c)(c+a)(a^2+b^2+c^2+ab+bc+ca)$

Maybe,It's very straight. But in my solution I didn't use it.

So what 's yours ? give some hints about your solution .....

A tighter inequality is this one $\frac{a^2+b^2+c^2}{a+b+c} \geq \sqrt{3}R$, which is the Garfulkel 's inequality.

I 'm sorry for my bad memory. It should be $\frac{a^2+b^2+c^2}{a+b+c}\leq \sqrt{3}R$.

Oh,along time and no one post the solution for this problem Here is then hint ,very neccessary: $sinAsin3A+sinBsin3B+sinCsin3C \le 0$ Please check it!

Then, the following better inequality holds $\frac{a^4+b^4+c^4}{a^3+b^3+c^3} \ge \sqrt{3}R$.

I start with the ineq: $\frac{a^4+b^4+c^4}{a^2+b^2+c^2} \ge 3R^2$ Which is equal with: $sinAsin3A+sinBsin3B+sinCsin3C \le 0$ And easy to prove $(\frac{a^5+b^5+c^5}{a^4+b^4+c^4})^2 \ge \frac{a^4+b^4+c^4}{a^2+b^2+c^2}$ Because $(a^5+b^5+c^5)(a^5+b^5+c^5)(a^2+b^2+c^2) \ge (a^4+b^4+c^4)^3$ (Holder) Vasc,Please proof $\frac{a^4+b^4+c^4}{a^3+b^3+c^3} \ge \sqrt{3}R$ I think it's tronger.

It is not very hard. We get the desired result by multiplying the inequalities $(a^4+b^4+c^4)(a^2+b^2+c^2) \ge (a^3+b^3+c^3)^2$ and $a^4+b^4+c^4 \ge 3R^2(a^2+b^2+c^2)$.

Oh yes. I am wrong.Thank you.

Can anybody please give a hint to that Jack Garfunkel's inequality? And also does anybody knows in what magazine (issue-year) was published? Thanks

hungkhtn wrote: I start with the ineq: $\frac{a^4+b^4+c^4}{a^2+b^2+c^2} \ge 3R^2$ Which is equal with: $sinAsin3A+sinBsin3B+sinCsin3C \le 0$ Yesterday I found in my collection the inequality $\frac{a^4+b^4+c^4}{a^2+b^2+c^2} \ge 3R^2$. The proof is the following. Since $R^2=\frac{a^2b^2c^2}{16S^2}=\frac {a^2b^2c^2}{2a^2b^2+2b^2c^2+2c^2a^2-a^4-b^4-c^4}$, using the notations $a^2=\frac{y+z}{2}$ etc., the inequality becomes as follows $xy(x-y)^2+yz(y-z)^2+zx(z-x)^2 \ge 0.$

GREAT PROOF, VASC

Also, I nearby found the following complementary inequalities for any triangle: (a) $(a+b+c)R \ge \sqrt{ab^3+bc^3+ca^3}$; (b) $(a+b+c)R \ge \sqrt{a^2b^2+b^2c^2+c^2a^2}$.

Why do you use this substitution? $\displaystyle a^2=\frac{y+z}{2} $ In my opinion $a^2,b^2,c^2$ are not sides of a triangle. $a^2+b^2 \geq \frac{(a+b)^2}{2} \geq \frac{c^2}{2}$ instead of $a^2+b^2 \geq c^2$ Where am I wrong, Vasc?

manlio, \[ a^2+b^2-c^2 = \frac{y+z+z+x-x-y}{2} = z \] and so on, so they are sidelengths of a triangle.

Manlio, the triangle is acute. hungkhtn wrote: Show that in acute triangle ABC we have: $\frac{a^5+b^5+c^5}{a^4+b^4+c^4} \ge \sqrt{3}R$

Show that in acute triangle $ABC$ , prove or disprove: $\frac{a^{n+1}+b^{n+1}+c^{n+1}}{a^n+b^n+c^n} \ge \sqrt{3}R$ $(n\in R^+,n\ge 2)$ $\frac{a^{n+2}+b^{n+2}+c^{n+2}}{a^n+b^n+c^n} \ge 3R^2$ $(n\in R^+,n\ge 2)$

We : use $ 2s\geq 3\sqrt{3}R, (*) $ $ \frac{\sum a^{n+1}}{\sum a^{n}}(CBS)\geq \frac{\sum a}{3}=\frac{2s}{3}\geq (*)=\sqrt{3}R $

We : use $ 2s\geq 3\sqrt{3}R, (*) $ $ \frac{\sum a^{n+2}}{\sum a^{n}}(CBS)\geq (\frac{\sum a}{3})^{2}\geq (*)=3R^{2} $

generalization : Let $ k,p\in N,k-p\geq 1 $ , then we : $ \frac{\sum a^{k}}{\sum a^{p}}\geq (\frac{a+b+c}{3})^{k-p}\geq (\sqrt{3}R)^{k-p} $

sinA+sinB+sinC≤ @nicusorz is right？

@Vasc your solving isn't right

wrong： $ 2s\geq 3\sqrt{3}R, (*) $

boi, this thread is 13 years old

Vasc wrote: Also, I nearby found the following complementary inequalities for any triangle: (b) $(a+b+c)R \ge \sqrt{a^2b^2+b^2c^2+c^2a^2}$. We have \[LHS^2 - RHS^2 = \left[12Rr+(p^2+5r^2-16Rr)\right] (4R^2+4Rr+3r^2-p^2)+4r(4R^2+3Rr+2r^2)(R-2r) \geqslant 0.\]

sqing wrote: Show that in acute triangle $ABC$ , prove or disprove: $\frac{a^{n+1}+b^{n+1}+c^{n+1}}{a^n+b^n+c^n} \ge \sqrt{3}R$ $(n\in R^+,n\ge 2)$ $\frac{a^{n+2}+b^{n+2}+c^{n+2}}{a^n+b^n+c^n} \ge 3R^2$ $(n\in R^+,n\ge 2)$ The first one holds for $\boxed{n\ge 3}$ . And they are both trivial because of the very nice inequality in non-obtuse-angled triangles which has already been mentioned, that is: \[\boxed{\frac {a^4 + b^4 + c^4}{a^2 + b^2 + c^2} \ge 3R^2}\ .\]This was also proven by other means here. Now, $n\mapsto \frac {a^{n+1} + b^{n+1} + c^{n+1}}{a^n + b^n + c^n}$ is decreasing and this is an immediate consequence of CS. And the base case for $n = 3$ follows again from CS together with the above-mentioned inequality. Similarly, $n\mapsto \frac {a^{n+2} + b^{n+2} + c^{n+2}}{a^n + b^n + c^n}$ is also a decreasing function on $n$ because $f(n)\ge f(n - 1)$ reduces to: \[ \begin{array}{lc} & \displaystyle \sum \left(b^{n-1}c^{n+2} + c^{n-1}b^{n+2}\right) \ge \sum \left(b^nc^{n+1} + c^nb^{n+1}\right) \\\\ \iff & \displaystyle \sum b^nc^n\left(\frac {c^2}b + \frac {b^2}c\right) \ge \sum b^nc^n\left(b + c\right), \end{array}\]which is obvious. As the base case coincides with the previous "boxed" inequality, there is nothing else to be done .

Nice.....

#14Feb 14, 2005, 9:05 pm hungkhtn wrote: I start with the ineq: $\frac{a^4+b^4+c^4}{a^2+b^2+c^2} \ge 3R^2$ Which is equal with: $sinAsin3A+sinBsin3B+sinCsin3C \le 0$ Yesterday I found in my collection the inequality $\frac{a^4+b^4+c^4}{a^2+b^2+c^2} \ge 3R^2$. The proof is the following. Since $R^2=\frac{a^2b^2c^2}{16S^2}=\frac {a^2b^2c^2}{2a^2b^2+2b^2c^2+2c^2a^2-a^4-b^4-c^4}$, using the notations $a^2=\frac{y+z}{2}$ etc., the inequality becomes as follows $xy(x-y)^2+yz(y-z)^2+zx(z-x)^2 \ge 0.$ right??????

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