We know a,b and c are positive and $a^2=b^2+c^2$ How we can conclude this inequality: $a^n>b^n+c^n$ , $n>2$
I tried Binomial Theorem but I can't prove this. Thanks
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1Hi matinhabi. Your question is very interesting, but it would be very useful if you had more details (what already do you have tried, your efforts, where you found this question, and so on). This will help a lot the community helping you to find your way toward the solution! – the_candyman Mar 14 '16 at 21:08
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My teacher in high school give it to me. – Mar 14 '16 at 21:10
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what was the topic your teacher was teaching you when he/she asked this question? – the_candyman Mar 14 '16 at 21:11
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He teaches Geometry – Mar 14 '16 at 21:12
3 Answers
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$$a^n=(a^2)^{\frac{n}{2}}=(b^2+c^2)^{\frac{n}{2}}>b^n+c^n$$
the inequality is equivalent to $(b^2+c^2)^n>(b^n+c^n)^2$ - this is not hard to prove by Newton's binomial formula
tong_nor
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$n>2\implies(b^2+c^2)^n>b^{2n}+n(b^2c^{2n-2}+b^{2n-2}c^2)+c^{2n}$ by Newton. now apply $x+y\ge 2\sqrt{xy}$ for the middle part and you're done! – tong_nor Mar 15 '16 at 13:07
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If $n=2m$ is even, then this is straightforward: $$ a^n=(a^2)^m=(b^2+c^2)^m> b^{2m}+c^{2m}=b^n+c^n$$
If $n>2$ is odd, then since $n-1$ is even it follows that $$ a^{n}=a\cdot a^{n-1}\geq a(b^{n-1}+c^{n-1})$$ using either equality if $n-1=2$, or what we have shown above if $n-1>2$.
Finally, $a^2=b^2+c^2$ with $a,b,c>0$ implies that $a>b$ and $a>c$, hence $$ a(b^{n-1}+c^{n-1})>b\cdot b^{n-1}+c\cdot c^{n-1}=b^n+c^n$$
carmichael561
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$a^2=b^2+c^2\iff 1=(\frac ba)^2+(\frac ca)^2$ and clearly $\frac ba <1$ and $\frac ca <1$. Hence $$1=(\frac ba)^2+(\frac ca)^2>(\frac ba)^3+(\frac ca)^3>(\frac ba)^4+(\frac ca)^4>......>(\frac ba)^n+(\frac ca)^n$$
Thus $$1>(\frac ba)^n+(\frac ca)^n\iff a^n>b^b+c^n$$
Piquito
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