most recent 30 from http://mathoverflow.net 2013-05-23T09:04:21Z http://mathoverflow.net/feeds/user/6180 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/26665/another-plausible-inequality/26773#26773 can_hang2007 2010-06-02T01:14:40Z 2010-06-02T01:25:49Z

<p>I think your inequality is false, dear miwalin. Please check the case when $a_1=b_2=\frac{\sqrt{3}}{2}$ and $a_2=b_1=-\frac{1}{2}.$ But I think <strong>it is true when $a_1,$ $a_2,$ $b_1,$ $b_2$ are nonnegative numbers.</strong></p> <p>Let me prove it in the case $a_1,$ $a_2,$ $b_1,$ $b_2$ are nonnegative real numbers. Write the inequality as $$\frac{(a_1^2+xa_2^2)(b_1^2+xb_2^2)}{(a_1b_1+xa_2b_2)^2} -1 \ge \left[ \frac{(1+x)+(1-x)(a_1b_1+a_2b_2)}{(1-x)+(1+x)(a_1b_1+a_2b_2)}\right]^2-1.$$ Since $$(a_1^2+xa_2^2)(b_1^2+xb_2^2)-(a_1b_1+xa_2b_2)^2=x(a_1^2b_2^2+a_2^2b_1^2-2a_1a_2b_1b_2)= x[(a_1^2+a_2^2)(b_1^2+b_2^2)-(a_1b_1+a_2b_2)^2]= x[1-(a_1b_1+a_2b_2)^2]$$ and $$\left[ \frac{(1+x)+(1-x)(a_1b_1+a_2b_2)}{(1-x)+(1+x)(a_1b_1+a_2b_2)}\right]^2-1=\frac{4x[1-(a_1b_1+a_2b_2)^2]}{[(1-x)+(1+x)(a_1b_1+a_2b_2)]^2},$$ the above inequality is equivalent to (notice that $x[1-(a_1b_1+a_2b_2)^2] \ge 0$) $$[(1-x)+(1+x)(a_1b_1+a_2b_2)]^2 \ge 4(a_1b_1+xa_2b_2)^2,$$ or $$(1-x)+(1+x)(a_1b_1+a_2b_2) \ge 2(a_1b_1+xa_2b_2),$$ or $$(1-x)(1-a_1b_1+a_2b_2) \ge 0,$$ which is obvious.</p>

http://mathoverflow.net/questions/25249/another-mixed-mean-inequality can_hang2007 2010-05-19T16:52:50Z 2010-05-20T03:14:28Z

<p>Let $a_1,$ $a_2,$ $\ldots,$ $a_n$ be positive real numbers. Prove that $$\sqrt{\frac{a_1^2+\left( \frac{a_1+a_2}{2}\right)^2+\cdots +\left(\frac{a_1+a_2+\cdots +a_n}{n}\right)^2}{n}} \le \frac{a_1+\sqrt{\frac{a_1^2+a_2^2}{2}}+\cdots+\sqrt{\frac{a_1^2+a_2^2+\cdots +a_n^2}{n}}}{n}.$$</p> <p>I have proved this inequality for $n=2$ and $n=3.$ But I still cannot prove it for the general case. Can somone help me?</p>

http://mathoverflow.net/questions/26665/another-plausible-inequality/26677#26677 can_hang2007 2010-06-02T02:39:04Z 2010-06-02T02:39:04Z

I think there is a mistake in your solution, dear Wadim Zudilin. The inequality $A \ge 0$ is false. Please check the case when $x=0,$ $u=\frac{1}{2}$ and $v=2.$ By the way, I think the inequality also does not hold for $a_2 b_2 \ge 0,$ but it holds for $a_1b_1+a_2b_2 \ge 0.$ We can modify my solution below a little to obtain this conclusion.

http://mathoverflow.net/questions/25249/another-mixed-mean-inequality/25319#25319 can_hang2007 2010-06-02T01:38:55Z 2010-06-02T01:38:55Z

Thanks Gjergji Zaimi for your reference.

http://mathoverflow.net/questions/26665/another-plausible-inequality/26773#26773 can_hang2007 2010-06-02T01:26:46Z 2010-06-02T01:26:46Z

After taking my values of $a_1,$ $a_2,$ $b_1,$ $b_2,$ the inequality is false for all $0&lt;x&lt;1.$