# If the sum of cubes of $a,b,c,d$ is $1$, then $\frac{1}{1-bcd}+\frac{1}{1-cda}+\frac{1}{1-dab}+\frac{1}{1-abc}\le \frac{16}{3}$

$a,b,c,d>0$ satisfying $a^3+b^3+c^3+d^3=1$. Prove $$\frac{1}{1-bcd}+\frac{1}{1-cda}+\frac{1}{1-dab}+\frac{1}{1-abc}\le \frac{16}{3}$$

I tried to go the normal way, by Cauchy-Schwarz, but that doesn't work. So I tried to incorporate this newly learned idea, since $a,b,c,d<1$ we can write the left as a power series: $$\sum_{n=0}^{\infty}(bcd)^n+(cda)^n+(dab)^n+(abc)^n$$ If we can show, $(bcd)^n+(cda)^n+(dab)^n+(abc)^n\ge (K(a^3+b^3+c^3+d^3))^n$ for some suitable constant $K$ we can finish. But I can't really do it. Can someone help me?

• For $n=1$, your last inequality doesn't look so promising. I guess same can be stated for $n>1$, but that needs verification. – Quang Hoang Aug 10 '14 at 14:45
• Yes, I see it now. But do you have any other proof of it? @QuangHoang – shadow10 Aug 10 '14 at 14:47
• Actually your reasoning is pretty good. I'm under the impression that all inequalities should be flipped. For example, if $a,b,c$ are close to zero, then the LHS is close to $1$. – Quang Hoang Aug 10 '14 at 14:51
• @QuangHoang Check $(a, b, c, d) \to (0, 0, 0, 1) \implies LHS \to 4$. – Macavity Aug 10 '14 at 16:49
• 0 down vote Yes, the inequality should be reversed. (This problem appears in the current issue of Mathematical Reflections, published by Titu Andreescu.) – PolyaPal Aug 14 '14 at 20:35

The AM-GM inequality yields

$$abcd \le \frac{1}{4 \sqrt[3]{4}} \iff abc \le \frac{1}{4 \sqrt[3]{4}d} \iff \frac{1}{1-abc} \le \frac{1}{1 - \frac{1}{4 \sqrt[3]{4} d}}.$$ It follows that $$\sum_{cyclic} \frac{1}{1-abc} \le \sum_{cyclic} \frac{4 \sqrt[3]{4} d}{4 \sqrt[3]{4} d -1}.$$ Therefore it is sufficient to show that $$\sum_{cyclic} \frac{4 \sqrt[3]{4} d}{4 \sqrt[3]{4} d -1} \le \frac{16}{3} \iff 4 - \sum_{cyclic} \frac{-1}{4 \sqrt[3]{4} d - 1} \le \frac{16}{3} \iff \sum_{cyclic} \frac{-1}{4 \sqrt[3]{4} d - 1} \ge \frac{-4}{3} \iff \sum_{cyclic} \frac{1}{1 - 4 \sqrt[3]{4}d} \ge \frac{-4}{3}.$$

Now using a form of the Cauchy-Schwarz inequality usually referred to as Titu's Lemma (after Titu Andreescu), we find that

$$\sum_{cyclic} \frac{1}{1 - 4 \sqrt[3]{4}d} \ge \frac{(1 + 1 +1 + 1)^2}{\sum_{cyclic} (1 - 4 \sqrt[3]{4}d)} = \frac{16}{4 - 4 \sqrt[3]{4}} (a + b + c + d).$$

Hence it suffices to show that

$$\frac{16}{4 - 4 \sqrt[3]{4}(a + b + c + d)} \ge \frac{-4}{3} \iff \frac{1}{1 - \sqrt[3]{4}(a + b + c + d)} \ge \frac{-1}{3} \iff 3 \ge -1 + \sqrt[3]{4} (a + b + c + d) \iff 4 \ge \sqrt[3]{4} (a + b + c + d) \iff \sqrt[3]{16} \ge a + b + c + d.$$ The last inequality follows from Hölder's inequality since $$(a^3 + b^3 + c^3 + d^3)(1 + 1 + 1 + 1)(1 + 1 + 1 + 1) \ge (a + b + c + d)^3 \iff \sqrt[3]{16} \ge a + b + c + d.$$ Therefore the conclusion follows and equality holds for $a = b = c = d = \frac{1}{\sqrt[3]{4}}.$

• This methods is not true.. – math110 Dec 26 '14 at 4:36
• I made two crucial LaTeX errors, which I have corrected. The proof is correct as it appears now. – PolyaPal Dec 27 '14 at 16:47
• I mean that: may$1-\sqrt[3]{4}(a+b+c+d)<0$,so $$\dfrac{1}{1-\sqrt[3]{4}(a+b+c+d)}\ge\dfrac{-1}{3}\Longleftrightarrow 3\ge -1+\sqrt[3]{4}(a+b+c+d)$$ is wrong! – math110 Dec 28 '14 at 6:19
• Thank you, math110. You are right. Several of the terms may be either positive or negative, thereby negating my use of basic properties of inequalities, including the Cauchy-Schwarz inequality. I'll keep trying to fix my proof. – PolyaPal Dec 30 '14 at 0:50
• OK. I checked with a very experienced problem proposer/solver, and it turns out that his solution is also flawed. He found that LHS $\le M$, but unfortunately $M \ge 16/3.$ This inequality is more subtle than we think. Maple calculations suggest that the proposed inequality is true however. – PolyaPal Dec 30 '14 at 22:46

Let $a+b+c+d=4u$, $ab+ac+bc+ad+bd+cd=6v^2$, $abc+abd+acd+bcd=4w^3$ and $abcd=t^4$.

Also let $abc=x$, $abd=y$, $acd=z$ and $bcd=p$.

Hence, we need to prove that: $$\sum_{cyc}\frac{1}{1-x}\leq\frac{16}{3}$$ or $$4-7\sum_{cyc}x+10(xy+xz+yz+xp+yp+zp)-13\sum_{cyc}xyz+16xyzp\geq0$$ or $$4(64u^3-72uv^2+12w^3)^4-28w^3(64u^3-72uv^2+12w^3)^3+$$ $$+60v^2t^4(64u^3-72uv^2+12w^3)^2-52ut^8(64u^3-72uv^2+12w^3)+16t^{12}\geq0$$ or $$64(16u^3-18uv^2+3w^3)^4-112w^3(16u^3-18uv^2+3w^3)^3+$$ $$+60v^2t^4(16u^3-18uv^2+3w^3)^2-13ut^8(16u^3-18uv^2+3w^3)+t^{12}\geq0.$$ We'll show that $$60v^2(16u^3-18uv^2+3w^3)^2-13ut^4(16u^3-18uv^2+3w^3)+t^{8}\geq$$ $$\geq48v^2(16u^3-18uv^2+3w^3)^2.$$ Indeed, we need to show that $$t^4\leq\frac{1}{2}(16u^3-18uv^2+3w^3)\left(13u-\sqrt{169u^2-48v^2}\right)$$ and since $t^4\leq v^4$, it's enough to prove that $$v^4\leq\frac{1}{2}(16u^3-18uv^2+3w^3)\left(13u-\sqrt{169u^2-48v^2}\right).$$ By Schur $w^3\geq4uv^2-3u^3$.

Thus, it remains to prove that $$v^4\leq\frac{1}{2}(7u^3-6uv^2)\left(13u-\sqrt{169u^2-48v^2}\right),$$ which is obvious because $u^2\geq v^2$.

Id est, it remains to prove that: $$64(16u^3-18uv^2+3w^3)^4-112w^3(16u^3-18uv^2+3w^3)^3+$$ $$+48v^2t^4(16u^3-18uv^2+3w^3)^2\geq0$$ or $$4(16u^3-18uv^2+3w^3)^2-7w^3(16u^3-18uv^2+3w^3)+3v^2t^4\geq0.$$ Now, $(a-b)^2(c-d)^2+(a-c)^2(b-d)^2+(a-d)^2(b-c)^2\geq0$ gives $t^4\geq4uw^3-3v^4$.

Hence, it remains to prove that $f(w^3)\geq0$, where $$f(w^3)=4(16u^3-18uv^2+3w^3)^2-7w^3(16u^3-18uv^2+3w^3)+3v^2(4uw^3-3v^4).$$ Now, by Schur $$f'(w^3)=30w^3+272u^3-294uv^2>$$ $$>22(4uv^2-3w^3)+272u^3-294uv^2=206u(u^2-v^2)\geq0,$$ which says that it's remains to prove $f(w^3)\geq0$ for a minimal value of $w^3$. Now, $a$, $b$, $c$ and $d$ are positive roots of the following equation $$(x-a)(x-b)(x-c)(x-d)=0$$ or $$x^4-4ux^3+6v^2x^2-4w^3x+t^4=0,$$ which says that the equation $$(x^4-4ux^3+6v^2x^2-4w^3x+t^4)'=0$$ or $$x^3-3ux^2+3v^2x-w^3=0$$ has three positive roots. Let $k$, $l$ and $m$ are the roots, which says that

$3u=k+l+m$, $3v^2=kl+km+lm$ and $w^3=klm$

and the equation $w^3=x^3-3ux^2+3v^2x$ has three positive roots.

Now, let $u$ and $v^2$ are constants. We see that $w^3$ gets a minimal value,

when the line $y=w^3$ is a tangent line to the graph of $y=x^3-3ux^2+3v^2x$,

which happens for equality case of two variables or else we must check $w^3\rightarrow0^+$.

Id est, it's enough to prove that $f(w^3)\geq0$ in the following cases.

1. $w^3\rightarrow0^+$.

Let $m\rightarrow0^+$ and $l=1$. We can assume that $l=1$ because $f(w^3)\geq0$ is homogeneous.

Thus, we need to prove here $$4(16u^3-18uv^2)^2-9v^6\geq0$$ or $$4\left(\frac{16(k+1)^3}{27}-2(k+1)k\right)^2-\frac{k^3}{3}\geq0$$ or $$1024k^6-768k^5-624k^4+2093k^3-624k^2-768k+1024\geq0,$$ which is obvious.

1. $l=m=1$.

In this case we obtain $$4\left(\frac{16(k+2)^3}{27}-2(k+2)(2k+1)+3k\right)^2-$$ $$-7k\left(\frac{16(k+2)^3}{27}-2(k+2)(2k+1)+3k\right)+(2k+1)\left(\frac{4(k+2)k}{3}-\frac{(2k+1)^2}{3}\right)\geq0$$ or $$(k-1)^2(1024k^4+512k^3-2064k^2-100k+1357)\geq0,$$ which is obviously true.

Done!