August 1, 2010

Problem 045 (Van Khea)

Filed under: Uncategorized — KKKVVV @ 11:10 pm

Let  a, b, c be positive real numbers and satisfying a\leq b\leq c. Prove that if  r, s, t>0 and r\geq s then we have

a^rb^sc^t+b^rc^sa^t+c^ra^sb^t\leq a^sb^{r+t}+b^sc^{t+r}+c^sa^{r+t}


We have a\leq b\leq c \Longrightarrow a^t\leq b^t\leq c^t ; a^r\leq b^r\leq c^r ; r, t>0

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

\displaystyle f''(x)=\frac{s(s-r)}{r^2}x^{\frac{s}{r}-2}<0

Let m=b^t, n=c^t, p=a^t; t>0\Longrightarrow m\leq n; p\leq n

From the inequality by Van Khea for m\leq n; p\leq n ; a^r\leq b^r\leq c^r and f''(x)<0 we have

m(c^r-b^r)f(a^r)-n(c^r-a^r)f(b^r)+p(b^r-a^r)f(c^r)\leq 0

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

\Longrightarrowa^rb^sc^t+b^rc^sa^t+c^ra^sb^t\leq a^sb^{r+t}+b^sc^{t+r}+c^sa^{r+t}

Example1: Let r=s=t=1 then we have 3abc\leq ab^2+bc^2+ca^2

Example2: Let r=2; s=t=1 then we have

abc(a+b+c)\leq ab^3+bc^3+ca^3

See also the inequality by Van Khea


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