July 28, 2010

(Van Khea):Problem 017

Filed under: the new math — KKKVVV @ 5:26 pm

Let a\geq b\geq c>0. Prove that:

\displaystyle \frac{a^{k}b}{c}+\frac{b^{k}c}{a}+\frac{c^{k}a}{b}\geq a^k+b^k+c^k ; \forall{k\geq 1}


Let k=1 so we need to prove that

\displaystyle \frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\geq a+b+c

From Cauchy’s inequality we have

\displaystyle \frac{ab}{c}+\frac{bc}{a}\geq 2\sqrt{\frac{ab^2c}{ca}}=2b

\displaystyle \frac{bc}{a}+\frac{ca}{b}\geq 2c

\displaystyle \frac{ca}{b}+\frac{ab}{c}\geq 2a

\displaystyle \Longrightarrow 2(\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b})\geq 2(a+b+c)

\displaystyle \Longrightarrow \frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\geq a+b+c

Therefor with k=1 the inequality is trued.

Now we suppose that k>1

So the inequality above became

\displaystyle \frac{a^kb}{c}-a^k+\frac{b^kc}{a}-b^k+\frac{c^ka}{b}-c^k\geq 0

From the conditional we had a\geq b\geq c>0 so we can rewrite

\displaystyle ca(a-b)c^k-bc(a-c)b^k+ab(b-c)a^k\geq 0

We have a\geq b\geq c\Longrightarrow ab\geq ca\geq bc

Let m=ca ; n=bc; p=ab\Longrightarrow m\geq n; p\geq n

Let f(x)=x^k ; x>0 ; k>1


From the inequality by van khea for a\geq b\geq c>0; m\geq n; p\geq n; f''(x)>0 we get

m(a-b)f(c)-n(a-c)f(b)+p(b-c)f(a)\geq 0

\displaystyle \Longrightarrow ca(a-b)c^k-bc(a-c)b^k+ab(b-c)a^k\geq 0

Therefor the proof is completed.

See also the inequality by Van Khea πŸ˜€


1 Comment »

RSS feed for comments on this post. TrackBack URI

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

Blog at

%d bloggers like this: