China Girls' Mathematical Olympiad

Solution 1. First, we will prove that, whenever there are two numbers among $a,b,c,d$ that are equal, the inequality holds. We may assume that $a=b$ and let $s=a+b+c+d$. Then we have $$\begin{array}{rl}& \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{9}{a+b+c+d}\\ & =\frac{2}{a}+\frac{c+d}{cd}+\frac{9}{s}=\frac{2}{a}+a^2(s-2a)+\frac{9}{s}\\ & =\frac{2}{a}-2a^3+\left(a^{2}s+\frac{9}{s}\right).\end{array}$$ We define the expression above as $f(s)$. Then we see that $f(s)$ reaches the minimum for $s=\frac{3}{a}$.

When $a\ge \frac{\sqrt{2}}{2}$, however, we have $$s=a+b+c+d\ge 2a+\frac{2}{a}\ge \frac{3}{a}.$$ At this time, $f(s)$ reaches the minimum for $s=2a+\frac{2}{a}$. We then have $$\begin{array}{rl}& \frac{2}{a}-2a^3+\left(a^{2}s+\frac{9}{s}\right)=\frac{2}{a}-2a^3+a^2\left(2a+\frac{2}{a}\right)+\frac{9}{s}\\ & =\frac{2}{a}+2a+\frac{9}{s}=s+\frac{9}{s}\\ & =\frac{7}{16}s+\frac{9}{16}s+\frac{9}{s}\ge \frac{7}{16}\times 4+2\sqrt{\frac{9}{16}s\cdot \frac{9}{s}}\\ & =\frac{7}{4}+\frac{9}{2}=\frac{25}{4}\left(\therefore s=2a+\frac{2}{a}\ge 4\right).\end{array}$$

When $0<a<\frac{\sqrt{2}}{2}$, we have

[The solution appears to be truncated here, but the main argument is present.]

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Alternative solution.

Solution 2. We prove it by using the adjustment method. We may assume that $a\le b\le c\le d$, and define $$f(a,b,c,d)=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{9}{a+b+c+d}.$$ Firstly, we will prove $$f(a,b,c,d)\ge f(\sqrt{ac},b,\sqrt{ac},d).\stackrel{◯}{2}$$ As a matter of fact, expression ② is equivalent to $$\begin{array}{rl}\frac{1}{a}+\frac{1}{c}+\frac{9}{a+b+c+d}& \ge \frac{1}{\sqrt{ac}}+\frac{1}{\sqrt{ac}}+\frac{9}{2\sqrt{ac}+b+d}\\ \iff \frac{(\sqrt{a}-\sqrt{c}{)}^2}{ac}& \ge \frac{9(\sqrt{a}-\sqrt{c}{)}^2}{(a+b+c+d)(2\sqrt{ac}+b+d)}\\ (\because (\sqrt{a}-\sqrt{c}{)}^2& \ge 0)\\ \iff (a+b+c+d)(2\sqrt{ac}+b+d)& \ge 9ac\\ (\because b+d& \ge 2\sqrt{bd}=\frac{2}{\sqrt{ac}})\\ \iff \left(a+c+\frac{2}{\sqrt{ac}}\right)\left(2\sqrt{ac}+\frac{2}{\sqrt{ac}}\right)& \ge 9ac.\end{array}$$ From $1=abcd\ge a\cdot a\cdot c\cdot c\Rightarrow ac\le 1\Rightarrow \frac{2}{\sqrt{ac}}\ge 2\sqrt{ac}$ and $a+c\ge 2\sqrt{ac}$, we have the following condition: The left-hand side of ③ $\ge \left(2\sqrt{ac}+\frac{2}{\sqrt{ac}}\right)\left(2\sqrt{ac}+\frac{2}{\sqrt{ac}}\right)$ $\ge 4\sqrt{ac}\cdot 4\sqrt{ac}=16ac>9ac$ is the right-hand side of ③. Therefore, ② holds, which means $f(a,b,c,d)$ (with $a\le b\le c\le d$) reaches its minimum if and only if $a=c$ (i.e. $a=b=c$). So we may assume that $(a,b,c,d)=\left(\frac{1}{t},\frac{1}{t},\frac{1}{t},t^3\right)$ ($t\ge 1$). Then we only need to prove that, for all $t\ge 1$, $$f\left(\frac{1}{t},\frac{1}{t},\frac{1}{t},t^3\right)\ge \frac{25}{4}.\stackrel{◯}{4}$$ We have $$\begin{array}{rl}f\left(\frac{1}{t},\frac{1}{t},\frac{1}{t},t^3\right)& \ge \frac{25}{4}\\ \iff 3t+\frac{1}{t^3}+\frac{9}{t^3+\frac{3}{t}}& \ge \frac{25}{4}\\ \iff 12t^8-25t^7+76t^4-75t^3+12& \ge 0\\ \iff (t-1{)}^2(12t^6-t^5-14t^4-27t^3+36t^2+24t+12)& \ge 0\\ \iff 12t^6-t^5-14t^4-27t^3+36t^2+24t+12& \ge 0\\ \iff (t-1)(12t^5+11t^4-3t^3-30t^2+6t+30)+42& \ge 0.\end{array}\stackrel{◯}{5}$$ Since $t\ge 1$, $12t^5+6t\ge 2\sqrt{12t^5\cdot 6t}=12\sqrt{2}{t}^3>3t^3$, and $$11t^4+30\ge 2\sqrt{11t^4\cdot 30}=2\sqrt{330}{t}^2>30t^2,$$ then $(t-1)(12t^5+11t^4-3t^3-30t^2+6t+30)+42>0$. Therefore, ⑤ holds, which justifies ④. The proof is complete.