51st IMO Shortlisted Problems

Observe that $$\begin{array}{c}4\left(a^3+b^3+c^3+d^3\right)-\left(a^4+b^4+c^4+d^4\right)=-\left((a-1{)}^4+(b-1{)}^4+(c-1{)}^4+(d-1{)}^4\right)\\ +6\left(a^2+b^2+c^2+d^2\right)-4(a+b+c+d)+4\\ =-\left((a-1{)}^4+(b-1{)}^4+(c-1{)}^4+(d-1{)}^4\right)+52\end{array}$$ Now, introducing $x=a-1,y=b-1,z=c-1,t=d-1$, we need to prove the inequalities $$16\ge x^4+y^4+z^4+t^4\ge 4$$ under the constraint $$\begin{array}{c}{x}^2+y^2+z^2+t^2=\left(a^2+b^2+c^2+d^2\right)-2(a+b+c+d)+4=4\end{array}\text{\hspace{1em}(1)}$$ (we will not use the value of $x+y+z+t$ though it can be found). Now the rightmost inequality in (1) follows from the power mean inequality: $$x^4+y^4+z^4+t^4\ge \frac{{\left(x^2+y^2+z^2+t^2\right)}^2}{4}=4$$ For the other one, expanding the brackets we note that $${\left(x^2+y^2+z^2+t^2\right)}^2=\left(x^4+y^4+z^4+t^4\right)+q$$ where $q$ is a nonnegative number, so $$x^4+y^4+z^4+t^4\le {\left(x^2+y^2+z^2+t^2\right)}^2=16$$ and we are done.

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Alternative solution.

First, we claim that $0\le a,b,c,d\le 3$. Actually, we have $$a+b+c=6-d,a^2+b^2+c^2=12-d^2$$ hence the power mean inequality $$a^2+b^2+c^2\ge \frac{(a+b+c{)}^2}{3}$$ rewrites as $$12-d^2\ge \frac{(6-d{)}^2}{3}⟺2d(d-3)\le 0$$ which implies the desired inequalities for $d$; since the conditions are symmetric, we also have the same estimate for the other variables. Now, to prove the rightmost inequality, we use the obvious inequality $x^2(x-2{)}^2\ge 0$ for each real $x$; this inequality rewrites as $4x^3-x^4\le 4x^2$. It follows that $$\left(4a^3-a^4\right)+\left(4b^3-b^4\right)+\left(4c^3-c^4\right)+\left(4d^3-d^4\right)\le 4\left(a^2+b^2+c^2+d^2\right)=48$$ as desired.

Now we prove the leftmost inequality in an analogous way. For each $x\in [0,3]$, we have $(x+1)(x-1{)}^2(x-3)\le 0$ which is equivalent to $4x^3-x^4\ge 2x^2+4x-3$. This implies that $\left(4a^3-a^4\right)+\left(4b^3-b^4\right)+\left(4c^3-c^4\right)+\left(4d^3-d^4\right)\ge 2\left(a^2+b^2+c^2+d^2\right)+4(a+b+c+d)-12=36$, as desired.

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Alternative solution.

First, expanding $48=4\left(a^2+b^2+c^2+d^2\right)$ and applying the AM-GM inequality, we have $$\begin{array}{rl}{a}^4+b^4+c^4+d^4+48& =\left(a^4+4a^2\right)+\left(b^4+4b^2\right)+\left(c^4+4c^2\right)+\left(d^4+4d^2\right)\\ & \ge 2\left(\sqrt{a^4\cdot 4a^2}+\sqrt{b^4\cdot 4b^2}+\sqrt{c^4\cdot 4c^2}+\sqrt{d^4\cdot 4d^2}\right)\\ & =4\left(\left|a^3\right|+\left|b^3\right|+\left|c^3\right|+\left|d^3\right|\right)\ge 4\left(a^3+b^3+c^3+d^3\right)\end{array}$$ which establishes the rightmost inequality. To prove the leftmost inequality, we first show that $a,b,c,d\in [0,3]$ as in the previous solution. Moreover, we can assume that $0\le a\le b\le c\le d$. Then we have $a+b\le b+c\le \frac{2}{3}(b+c+d)\le \frac{2}{3}\cdot 6=4$. Next, we show that $4b-b^2\le 4c-c^2$. Actually, this inequality rewrites as $(c-b)(b+c-4)\le 0$, which follows from the previous estimate. The inequality $4a-a^2\le 4b-b^2$ can be proved analogously. Further, the inequalities $a\le b\le c$ together with $4a-a^2\le 4b-b^2\le 4c-c^2$ allow us to apply the Chebyshev inequality obtaining $$\begin{array}{rl}{a}^2\left(4a-a^2\right)+b^2\left(4b-b^2\right)+c^2\left(4c-c^2\right)& \ge \frac{1}{3}\left(a^2+b^2+c^2\right)\left(4(a+b+c)-\left(a^2+b^2+c^2\right)\right)\\ & =\frac{\left(12-d^2\right)\left(4(6-d)-\left(12-d^2\right)\right)}{3}.\end{array}$$ This implies that $$\begin{array}{rl}\left(4a^3-a^4\right)& +\left(4b^3-b^4\right)+\left(4c^3-c^4\right)+\left(4d^3-d^4\right)\ge \frac{\left(12-d^2\right)\left(d^2-4d+12\right)}{3}+4d^3-d^4\\ & =\frac{144-48d+16d^3-4d^4}{3}=36+\frac{4}{3}(3-d)(d-1)\left(d^2-3\right)\end{array}\text{\hspace{1em}(2)}$$ Finally, we have $d^2\ge \frac{1}{4}\left(a^2+b^2+c^2+d^2\right)=3$ (which implies $d>1$ ); so, the expression $\frac{4}{3}(3-d)(d-1)\left(d^2-3\right)$ in the right-hand part of (2) is nonnegative, and the desired inequality is proved.