Another plausible inequality. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T23:27:05Z http://mathoverflow.net/feeds/question/26665 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/26665/another-plausible-inequality Another plausible inequality. Sunni 2010-06-01T04:37:39Z 2010-06-02T03:36:54Z <p>I come across the following problem in my study. </p> <p>Consider in the real field. Let $0\le x\le1$, $a_1^2+a_2^2=b_1^2+b_2^2=1$.Is it true</p> <p>$(a_1b_1+xa_2b_2)^2\le\left(\frac{(1-x)+(1+x)(a_1b_1+a_2b_2)}{(1+x)+(1-x)(a_1b_1+a_2b_2)}\right)^{2}(a_1^2+xa_{2}^{2})(b_1^2+xb_{2}^{2})$?</p> http://mathoverflow.net/questions/26665/another-plausible-inequality/26677#26677 Answer by Wadim Zudilin for Another plausible inequality. Wadim Zudilin 2010-06-01T07:31:02Z 2010-06-02T03:36:54Z <p>As Can Hang points out in his response, the inequality does not hold in general. Thanks to his comment to my own post, I stand corrected and claim the inequality is valid at least for the case $$a_1b_1+xa_2b_2\ge0 \qquad(*)$$ (and this seems to be a necessary condition as well).</p> <p>Let me do some standard things. First let $$a_1=\frac{1-u^2}{1+u^2}, \quad a_2=\frac{2u}{1+u^2}, \quad b_1=\frac{1-v^2}{1+v^2}, \quad b_2=\frac{2v}{1+v^2}$$ where $uv\ge0$. Substitution reduces the inequality to the following one: $$((1-u^2)(1-v^2)+4xuv)^2 \le\biggl(\frac{(uv+1)^2-x(u-v)^2}{(uv+1)^2+x(u-v)^2}\biggr)^2 ((1-u^2)^2+4xu^2)((1-v^2)^2+4xv^2). \qquad{(1)}$$ Now introduce the notation $$A=(1-u^2)(1-v^2)+4xuv, \quad B=(uv+1)^2, \quad C=x(u-v)^2$$ and note that $A,B,C$ are nonnegative; the inequality $A\ge0$ is equivalent to the above condition $(*)$. In addition, $$A\le B-C \qquad{(2)}$$ because $$B-C-A=(1-x)(u+v)^2\ge0.$$ In the new notation the inequality (1) can be written more compact: $$A^2(B+C)^2\le(B-C)^2(A^2+4BC)$$ which after straightforward reduction becomes $$A^2\le(B-C)^2,$$ while the latter follows from (2).</p> http://mathoverflow.net/questions/26665/another-plausible-inequality/26773#26773 Answer by can_hang2007 for Another plausible inequality. 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>