INEQUALITY'S BLOG

August 10, 2010

Problem 048

Filed under: Uncategorized — KKKVVV @ 1:31 pm

(Van Khea): Let a, b, c>0. Prove that:

\displaystyle \biggl(\frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab}\biggl)(a^3+b^3+c^3)\geq (ab+bc+ca)^2

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Problem 047

Filed under: Uncategorized — KKKVVV @ 1:24 pm

(Van Khea): Let a, b, c>0; a^2+b^2+c^2=1. Prove that

\displaystyle \frac{a^{\frac{2}{3}}}{(b+c)^{2}}+\frac{b^{\frac{2}{3}}}{(c+a)^2}+\frac{c^{\frac{2}{3}}}{(a+b)^2}\geq \frac{3^{\frac{2}{3}}}{4(a^4+b^4+c^4)}

August 1, 2010

Problem 046

Filed under: Uncategorized — KKKVVV @ 11:32 pm

Let a, b, c\geq 0 ; a\leq b\leq c and s\geq r\geq 0; s\geq 2. Prove that:

\displaystyle \biggl(\frac{a^rb^s+b^rc^s+c^ra^r}{3}\biggl)\biggl(\frac{a^{s-r}+b^{s-r}+c^{s-r}}{3}\biggl)\geq \biggl(\frac{ab+bc+ca}{3}\biggl)^s

 

Van Khea          

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}

Proof

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

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

Problem 043

Filed under: Uncategorized — KKKVVV @ 1:39 pm

 Prove that for c\geq b\geq a>0 we have:

\displaystyle \frac{a}{a+c}+\frac{b}{b+a}+\frac{c}{c+b}\geq \frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a} 

 

Van khea

Problem 042

Filed under: Uncategorized — KKKVVV @ 8:09 am

Let f:R\longrightarrow R_{0}^{+} and f''>0. For all 0<a\leq b\leq c ; a, b, c\in I ; Prove that:

c^3a^2f(a)+a^3b^2f(b)+b^3c^2f(c)\geq abc(caf(a)+abf(b)+bcf(c))

 

Van Khea

Problem 041

Filed under: Uncategorized — KKKVVV @ 7:42 am

Let f;R\longrightarrow R_{0}^{+} and f''>0 . For a, b, c be positive real numbers and satisfying a\leq b\leq c. Prove that:

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

Van Khea

Problem 040

Filed under: Uncategorized — KKKVVV @ 7:26 am

Let f:R\longrightarrow R_{0}^{+} and f''>0. For a, b, c be positive real numbers and satisfying a\leq b\leq c. Prove that:

\displaystyle \frac{a^2f(b)}{b+c}+\frac{b^2f(c)}{c+a}+\frac{c^2f(a)}{a+b}\geq \frac{abf(c)}{c+a}+\frac{bcf(a)}{a+b}+\frac{caf(b)}{b+c}

 

Van Khea    

Problem 039

Filed under: Uncategorized — KKKVVV @ 6:52 am

Let f:R\longrightarrow R_{0}^{+} and f''>0. For a, b, c be positive real numbers and satisfying a\leq b\leq c; ab+bc+ca=1 ; Prove that:

a^2f(b)+b^2f(c)+c^2f(a)\geq f(3abc)

Proof

From the problem 038 we have

a^2f(b)+b^2f(c)+c^2f(a)\geq abf(c)+bcf(a)+caf(b)

From Jensen’s inequality for ab+bc+ca=1 we have

abf(c)+bcf(a)+caf(b)\geq f(abc+bca+cab)=f(3abc)

\Longrightarrowa^2f(b)+b^2f(c)+c^2f(a)\geq f(3abc)

Therefore the proof is completed.

 

Van Khea   

Problem 038

Filed under: Uncategorized — KKKVVV @ 6:44 am

Let f:R\longrightarrow R_{0}^{+} and f''>0. For all 0<a\leq b\leq c ; a, b, c\in I ; Prove that:

a^2f(b)+b^2f(c)+c^2f(a)\geq abf(c)+bcf(a)+caf(b)

Proof

Let m=c, n=a ; p=b but we have a\leq b\leq c\Longrightarrow m\geq n; p\geq n

From the inequality by van khea for m\geq n; p\geq n ; a\leq b\leq c and f''>0 we have

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

\Longrightarrow c(c-b)f(a)-a(c-a)f(b)+b(b-a)f(c)\geq 0

\Longrightarrow a^2f(b)+b^2f(c)+c^2f(a)\geq abf(c)+bcf(a)+caf(b)

Therefor the proof is completed.

 

Van Khea    

 

See also the inequality by Van Khea

Problem 037

Filed under: Uncategorized — KKKVVV @ 6:36 am

Let f:R\longrightarrow R_{0}^{+} and f''>0. For all a\geq b\geq c>0; a, b, c\in I, Prove that:

ab^2f(a)+bc^2f(b)+ca^2f(c)\geq abc(f(a)+f(b)+f(c))

 

Van khea     

Problem 036

Filed under: Uncategorized — KKKVVV @ 1:31 am

Let a\geq b\geq c>0; a+b+c=abc. Prove that:

\displaystyle \frac{ab^2}{\sqrt{b+c}}+\frac{bc^2}{\sqrt{c+a}}+\frac{ca^2}{\sqrt{a+b}}\geq \frac{3\sqrt{3}}{\sqrt{2}}\sqrt{a+b+c}

 

Van Khea

Problem 035

Filed under: Uncategorized — KKKVVV @ 1:25 am

Let a\geq b\geq c>0; a+b+c=1. Prove that:

\displaystyle \frac{ab^2}{\sqrt{b+c}}+\frac{bc^2}{\sqrt{c+a}}+\frac{ca^2}{\sqrt{a+b}}\geq \frac{3\sqrt{3}abc}{\sqrt{2}}

 

Van Khea

Problem 034

Filed under: Uncategorized — KKKVVV @ 1:12 am

Let f:R\longrightarrow R_{0}^{+} and f''>0. Prove that for a\geq b\geq c>0; a,b,c\in I; we have

\displaystyle \frac{ab^2f(a)+bc^2f(b)+ca^2f(c)}{3}\geq abcf\biggl(\frac{a+b+c}{3}\biggl)

 

Van Khea   

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