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Prove that $$4(a+b+c)\ge 3(abc+3); \forall a,b,c> 0: a^2+b^2+c^2=a^2b^2+b^2c^2+c^2a^2.$$ I've tried $pqr$ method.

Let $p=a+b+c;q=ab+bc+ca;r=abc.$ The condition gives $$r=\frac{q^2+2q-p^2}{2p}$$

Now, we rewrite the inequality as $$4p\ge 9+3\cdot \frac{q^2+2q-p^2}{2p}$$which is reduced to $$11p^2\ge 18p+3q^2+6q $$ I didn't know how to continue. Could you please help me? Thank you

Vezen BU
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Hello world
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1 Answers1

1

We use the pqr method.

Let $p = a + b + c, q = ab + bc + ca, r = abc$. The condition is written as $$p^2 - 2q = q^2 - 2pr,$$ or $$r = \frac{q^2 + 2q - p^2}{2p}. \tag{1}$$

We need to prove that $$4p \ge 3(r + 3),$$ or (using (1)) $$11p^2 - 18p - 3q^2 - 6q \ge 0. \tag{2}$$

Using (1) and $q^2 \ge 3pr$, we have $q^2 \ge 3p \cdot \frac{q^2 + 2q - p^2}{2p}$ which results in $$p \ge \sqrt{\frac{q^2 + 6q}{3}}. \tag{3}$$

Using (1) and $r \ge \frac{4pq - p^3}{9}$ (three degree Schur inequality), we have $\frac{q^2 + 2q - p^2}{2p} \ge \frac{4pq - p^3}{9}$ which results in $$2p^4 - (8q + 9)p^2 + 9q^2 + 18q \ge 0. \tag{4}$$

We split into two cases.

Case 1. $q \ge 3$

Using (2) and (3), it suffices to prove that $$11\left(\sqrt{\frac{q^2 + 6q}{3}}\right)^2 - 18\cdot \sqrt{\frac{q^2 + 6q}{3}} - 3q^2 - 6q \ge 0$$ which is true. (Note: We use the fact that $x \mapsto 11x^2 - 18x$ is strictly increasing on $x \ge 2$.)

Case 2. $q < 3$

From (3) and (4), we have $$p \ge \sqrt{2q + \frac94 + \frac14\sqrt{81 - 8q^2}}. \tag{5}$$

Using (2) and (5), it suffices to prove that $$11\left(\sqrt{2q + \frac94 + \frac14\sqrt{81 - 8q^2}}\right)^2 - 18\sqrt{2q + \frac94 + \frac14\sqrt{81 - 8q^2}} - 3q^2 - 6q \ge 0$$ which is true. (Note: We use the fact that $x \mapsto 11x^2 - 18x$ is strictly increasing on $x \ge 2$.)

We are done.

River Li
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  • Thank you. I didn't expect it is complicated but this inequality seems hard. Btw, have you checked another question? https://math.stackexchange.com/questions/4877684/how-to-prove-abc-ge-3-with-the-strange-condition – Hello world Mar 13 '24 at 23:50
  • I think this pqr is kind of easy one (but complicated) (Note: Easy vs Difficult; Simple vs Complicated.). In this answer, the idea is described as follows. $$\left. \begin{array}{r} r = \frac{q^2 + 2q - p^2}{2p} \ q^2 \ge 3pr \ r \ge \frac{4pq - p^3}{9}\ p, q, r > 0 \end{array} \right} \implies 4p \ge 3(r + 3).$$ – River Li Mar 14 '24 at 00:03
  • @Helloworld By the way, the Buffalo Way (BW) kills it (but sure it is complicated). To prove $4(a + b + c)Q - 3(abc + Q^{3/2})$ where $Q := \frac{a^2b^2 + b^2c^2 + c^2a^2}{a^2 + b^2 + c^2}$. – River Li Mar 14 '24 at 00:40