INEQUALITY'S BLOG

August 1, 2010

Problem 044

Filed under: Uncategorized — KKKVVV @ 10:48 pm

Let a, b, c be positive real numbers and satisfying a\leq b\leq c. Prove that:

  • If  r\geq s\geq 0 then we have

a^sb^r+b^sc^r+c^sa^r\geq a^rb^s+b^rc^s+c^ra^s

  • If  s\geq r\geq 0 then we have

 a^sb^r+b^sc^r+c^sa^r\leq a^rb^s+b^rc^s+c^ra^s

Proof

We have a\leq b\leq c\Longrightarrow a^s\leq b^s\leq c^s ; s>0

Let \displaystyle f(x)=x^{\frac{r}{s}} ; x>0

\displaystyle f''(x)=\frac{r}{s}(\frac{r}{s}-1)x^{\frac{r}{s}-2}

  • If r\geq s then f''(x)>0 So from the inequality by VanKhea for

m=n=p ; a^s\leq b^s\leq c^s and f''(x)>0 we get

(c^s-b^s)f(a^s)-(c^s-a^s)f(b^s)+(b^s-a^s)f(c^s)\geq 0

\Longrightarrow (c^s-b^s)a^r-(c^s-a^s)b^r+(b^s-a^s)c^r\geq 0

\Longrightarrow a^sb^r+b^sc^r+c^sa^r\geq a^rb^s+b^rc^s+c^ra^s

  • If r\leq s then we have f''(x)<0 So from the inequality by Van Khea for m=n=p ; a^s\leq b^s\leq c^s and f''(x)<0 we get:

(c^s-b^s)f(a^s)-(c^s-a^s)f(b^s)+(b^s-a^s)f(c^s)\leq 0

\Longrightarrow (c^s-b^s)a^r-(c^s-a^s)b^r+(b^s-a^s)c^r\leq 0

\Longrightarrow a^sb^r+b^sc^r+c^sa^r\leq a^rb^s+b^rc^s+c^ra^s

 

Van Khea     

See also the inequality by van khea

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