I got this problem I tried several time to solve it by many inequalities but I got stuk. My question is how I get the minimum value of $$ \frac{a^{10}+ b^{10}}{a^{7}+ b^{7}} +\frac{b^{10}+ c^{10}}{b^{7}+ c^{7}} +\frac{c^{10}+ a^{10}}{c^{7}+ a^{7}} $$ if you know that $a, b, c \in (0, \infty) $ and $ a+b+c=1$? Any hint?
Asked
Active
Viewed 187 times
5
-
1It's probably a stretch, but you could try using Lagrange multipliers. – user170231 Oct 06 '15 at 17:13
-
@user170231 sorry but could you add more explanation – user260699 Oct 06 '15 at 18:41
-
The idea would be to implement the method used here, but I suspect that might make things complicated without the help of a CAS. – user170231 Oct 06 '15 at 18:56
3 Answers
9
First note that $$(a^3-b^3)(a^7-b^7)\ge 0\tag{1}$$ so $$2(a^{10}+b^{10})\ge (a^7+b^7)(a^3+b^3).\tag{2}$$ It follows that $$\frac{a^{10}+b^{10}}{a^7+b^7}+\frac{b^{10}+c^{10}}{b^7+c^7}+\frac{c^{10}+a^{10}}{c^7+a^7}\ge a^3+b^3+c^3.$$ Note that the equality happens when $a=b=c$. Next, we show that $a^3+b^3+c^3$ attains its minimum also when $a=b=c$. This can be seen by showing $$3(a^3+b^3+c^3)\ge (a+b+c)(a^2+b^2+c^2)\ge (a+b+c)\frac{(a+b+c)^2}{3}\tag{3},$$ using Rearrangement Inequality, which can be proved as in (1) $\implies$ (2).
Or we can use Cauchy-Schwarz inequality $$(a+b+c)(a^3+b^3+c^3)\ge (a^2+b^2+c^2)^2\ge \frac{(a+b+c)^4}{3^2}.$$
Quang Hoang
- 15,854
-
Is there a symmetry argument one could make to go directly to the extreme being at a=b=c? – Jerry Guern Oct 07 '15 at 00:57
3
Another (though similar) way... by Chebyshev's inequality applied twice:
$$ \sum_{cyc} \frac{a^{10}+b^{10}}{a^7+b^7} \ge \sum_{cyc}\frac{a^3+b^3}2 = \sum_{cyc}a^3\ge \frac19(a+b+c)^3 =\frac19$$ with equality when $a=b=c=\frac13$, so that's the min.
Macavity
- 46,381
1
$2(a^{10}+b^{10})\ge (a^7+b^7)(a^3+b^3)$ with chebishev inequality, then $$\frac{a^{10}+b^{10}}{a^7+b^7}+\frac{b^{10}+c^{10}}{b^7+c^7}+\frac{a^{10}+c^{10}}{a^7+c^7} \ge a^3+b^3+c^3 \ge 3 (\frac{a+b+c}{3})^3=\frac{1}{9}$$ because with power mean we have $ (\frac{a^3+b^3+c^3}{3})^\frac{1}{3} \ge (\frac{a+b+c}{3})$
yavuz
- 463