Thứ Ba, 16 tháng 4, 2013

Inequality from Czech-Polish-Slovak Match 2012

To Sayan Das.

Problem. Let $a,\,b,\,c,\,d$ be positive real numbers satisfying $a^2+b^2+c^2+d^2=10$ and $abcd=4.$ Determine the maximum value of the expression: $$P=(a+c)(b+d).$$
Solution. Without loss of generality, we may assume that $ac \ge bd.$ From this, we have $a^2c^2 \ge abcd=4$ and hence  $ac \ge 2.$

Now, using the AM-GM inequality, we have $$10=(a^2+c^2)+(b^2+d^2) \ge 2ac+2bd=2ac+\frac{8}{ac}.$$ Solving the inequality $2ac+\frac{8}{ac} \le 10,$ we get $1 \ge ac \le 4.$ Combining with the result $ac \ge 2$ obtained above, we get $2 \le ac \le 4.$

Setting $h=a^2+c^2$ and $t=ac$ $(h \ge 2t,\, 2 \le t \le 4).$ Under this substitution, we have $$(a+c)^2=h+2t,\quad (b+d)^2=b^2+d^2+2bd=10-h+\frac{8}{t}.$$ Therefore, $$P^2=(a+c)^2(b+d)^2=(h+2t)\left( 10-h+\frac{8}{t}\right) =\left(t+5+\frac{4}{t}\right)^2 -\left(h+t-5-\frac{4}{t}\right)^2 .$$ Let's consider two cases:
  • Case $3t^2 -5t -4 \ge 0.$ In this case, we have $$h+t-5-\frac{4}{t} \ge 3t-5-\frac{4}{t}=\frac{3t^2-5t-4}{t} \ge 0.$$Therefore, $$\begin{aligned} P^2& \le \left(t+5+\frac{4}{t}\right)^2 -\left(3t-5-\frac{4}{t}\right)^2=8\left[t(5-t)+4\right] \\ &\le 8 \left\{ \left[ \frac{t+(5-t)}{2}\right]^2+4\right\}=82.\end{aligned}$$ Equality holds when $h=2t,\, t=\frac{5}{2}, a^2+b^2+c^2+d^2=10$ and $abcd=4.$ This can be attained for example when $$a=c=\sqrt{\frac{5}{2}},\quad b=\frac{\sqrt{205}+3\sqrt{5}}{10}, \quad d=\frac{\sqrt{205}-3\sqrt{5}}{10}.$$
  • Case $3t^2-5t-4<0.$ In this case, we have $2 \le t <\frac{5+\sqrt{73}}{6}.$ Since the function $f(t)=t+\frac{4}{t}+5$ is increasing for $t \ge 2,$ we get $$f(t) <f\left(\frac{5+\sqrt{73}}{6}\right)=\frac{10+2\sqrt{73}}{3}.$$ And thus, it follows that $$P^2 \le \left(t+5+\frac{4}{t}\right)^2 <\left(\frac{10+2\sqrt{73}}{3}\right)^2<82.$$
From the two cases considered above, it is easy to see that $\max P=\sqrt{82}.$

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