Show that $(a + b)^4 + (a + c)^4 + (a + d)^4 + (b + c)^4 + (b + d)^4 + (c + d)^4 \le 6$ for $a^2 + b^2 + c^2 + d^2 = 1$.

Question

For $a, b, c, d \in \Bbb R$ such that $a^2 + b^2 + c^2 + d^2 = 1$, show that $$(a + b)^4 + (a + c)^4 + (a + d)^4 + (b + c)^4 + (b + d)^4 + (c + d)^4 \le 6.$$

The answer uses the mysterious identity $$(a + b)^4 + (a + c)^4 + (a + d)^4 + (b + c)^4 + (b + d)^4 + (c + d)^4 + (a - b)^4 + (a - c)^4 + (a - d)^4 + (b - c)^4 + (b - d)^4 + (c - d)^4 = 6(a^2 + b^2 + c^2 + d^2)^2.$$

But this just comes out of the blue. Are there any other solutions?

Edit (for those who say this is lacking context): This comes from the beginning of a book named Algebraic Inequalities (not in English). After the preface to the first chapter, where the author introduced that many inequalities can be proved by using the non-negativity of $x^2$, this is the first example (without developing any other methods yet). Then the author notes: the reader is supposed to memorize identities like that.

Answer 1

Note that $(a+b)^4+(a-b)^4=2a^4+12a^2b^2+2b^2$. Hence $\begin{align}(a + b)^4 &+ (a + c)^4 + (a + d)^4 + (b + c)^4 + (b + d)^4 + (c + d)^4\\ &+ (a - b)^4 + (a - c)^4 + (a - d)^4 + (b - c)^4 + (b - d)^4 + (c - d)^4\\ &=6(a^4+b^4+c^4+d^4)+12(a^2b^2+b^2c^2+c^2d^2+d^2a^2+a^2c^2+b^2d^2)\\ &= 6(a^2 + b^2 + c^2 + d^2)^2.\end{align}$

Answer 2

I give two solutions.

Solution 1:

After homogenization, it suffices to prove that $$(a + b)^4 + (a + c)^4 + (a + d)^4 + (b + c)^4 + (b + d)^4 + (c + d)^4 \le 6(a^2+b^2+c^2+d^2)^2.$$

The Buffalo Way works. WLOG, assume that $d \le c \le b \le a$. Let $c = d + s, b = d + s + t, a = d + s + t + r$ for $s, t, r\ge 0$. We have \begin{align} \mathrm{RHS} - \mathrm{LHS} &= 3 r^4+4 r^3 s+8 r^3 t+6 r^2 s^2+12 r^2 s t+12 r^2 t^2+4 r s^3\\ &\quad +12 r s^2 t +12 r s t^2+8 r t^3+3 s^4+8 s^3 t+12 s^2 t^2+8 s t^3+4 t^4. \end{align} We are done.

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Solution 2:

We can use SOS (Sum of Squares) method without the identity, with the help of computer.

After homogenization, it suffices to prove that $$(a + b)^4 + (a + c)^4 + (a + d)^4 + (b + c)^4 + (b + d)^4 + (c + d)^4 \le 6(a^2+b^2+c^2+d^2)^2.$$

We have \begin{align} \mathrm{RHS} - \mathrm{LHS} &= (-a b+a c+b d-c d)^2+\frac{1}{3} (a b+a c-2 a d-2 b c+b d+c d)^2 \\ &\quad +\frac{1}{2} (-a^2+2 a d+b^2-2 b c+c^2-d^2)^2\\ &\quad +(a b-a c-b^2+b d+c^2-c d)^2+(a^2-a b-a c+b d+c d-d^2)^2\\ &\quad +\frac{1}{6} (-3 a^2+2 a b+2 a c+2 a d-3 b^2+2 b c+2 b d-3 c^2+2 c d-3 d^2)^2. \end{align} We are done.