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Let $a,b,c$ be positive real numbers such, prove $$\frac{1}{7a^2+bc}+\frac{1}{7b^2+ca}+\frac{1}{7c^2+ab} \ge \frac{9}{8(ab+bc+ca)}$$ I saw it here. I try to full expanding but it's very complicated. Also, Cauchy Schwartz $$\sum_{cyc} \frac{1}{7a^2+bc} \ge \frac{9}{7(a^2+b^2+c^2)+(ab+bc+ca)}$$ leads us to a wrong inequality. Then I try $2bc \le b^2+c^2$ without success. So far, I haven't had any good idea yet. Can someone help me with this problem? Thank you.

My expanding $$\sum_{sym} \left(56a^4b^2+56a^4c^2+49a^4bc-\frac{49}{2}(ab)^3-\frac{49}{2}(ac)^3+456a^3b^2c+456a^3bc^2-1024a^2b^2c^2\right) \ge 0$$ it seems true as Murihead but I don't know how to use it.

30 Anh Ti 711
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  • We may use the pqr method. You need some calculations to find the pqr expression. – River Li Jan 27 '24 at 15:24
  • @RiverLi yes, it's obvious that f(r) is concave, so we just need to prove inequality for a=b, I just required any nice solution for this problem. – 30 Anh Ti 711 Jan 28 '24 at 01:20
  • Once you get the pqr expression, it is not difficult (I think it is easy actually since it is quadratic and concave in $r$). By the way, what you wrote is uvw rather than pqr. In pqr method, we don't check inequality for e.g. $a=b$ (uvw does). – River Li Jan 28 '24 at 01:35
  • @RiverLi What’s the difference between $pqr$ and $uvw$ methods? I thought those were the same since all the articles on the topic I’ve found stated that. (Except one comment on MSE that said “$pqr$-method = $uvw$-method $\neq$ $uvw$-technique”.) – Aig Feb 03 '24 at 15:55
  • @Aig I distinguish the pqr method and the uvw method. In my opinion, (i) The pqr method uses some relations among $p, q, r$ which are not necessarily optimal, while the uvw method uses the relations among $u, v, w$ in an optimal way. For example, in the uvw method, we fix $u, v^2$ and investigate when $w^3$ is minimal or maximal; while in the pqr method, we fix $p, q$ and find the inequalities about $r$ such as $r \ge \frac{4pq - p^3}{9}$, $r \le \frac{q^2}{3p}$ which are not the optimal bounds for $r$. (See next comment) – River Li Feb 04 '24 at 00:01
  • @Aig (ii) Usually, we use the uvw method by invoking the uvw theorem to say something like "we only need to check the case where two of $a,b,c$ are equal, or the case where one of $a, b, c$ are zero". The pqr method does not invoke any 'pqr' theorem, but deal with $p, q, r$ directly. (see next comment) – River Li Feb 04 '24 at 00:16
  • @Aig (iii) I think that the uvw method cannot be used for the following problem while the pqr method works. Problem: Let $a, b, c \ge 0$ such that $abc[(a-b)(b-c)(c-a)]^2 = 1$. Prove that $$a+b+c \ge \sqrt[9]{\frac{531441}{16}}.$$ In this problem, "the case where two of $a,b,c$ are equal, or the case where one of $a, b, c$ are zero" are excluded. – River Li Feb 04 '24 at 00:17
  • @RiverLi I examined your solution on MSE for this problem, indeed it looks very unlike the standard uvw-method. In some cases you were doing even the opposite of what I would expect from uvw. Thanks for clarification! – Aig Feb 04 '24 at 00:55
  • @Aig The word 'standard' is appropriate. I think that the standard pqr method is simply a substitution + the well-known inequalities such as Schur's inequality $p^3 + 9r - 4pq \ge 0$ ; The standard uvw invokes the uvw theorems to check the case where two of $a,b,c$ are equal, or the case where one of $a,b,c$ are zero. – River Li Feb 04 '24 at 02:16
  • @RiverLi I tried to prove the OP’s inequality by the $pqr$-method. I got the $pqr$-expression, then with the help of some inequalities (namely, $pq\geq 9r$, $p^2q^2+12r^2\geq 4p^3r+pqr$ and $p^3r\geq q^3$) I reduced it to the inequality $1344r^2\geq 0$ which is obviously true. But this big gap between $1344r^2$ and $0$ bothers me. Does this look realistic? – Aig Feb 04 '24 at 06:39
  • @Aig The equality case is $a = b = c = 1$ (so $r = 1$). The pqr expression is $-4608,{r}^{2}+ \left( -63,{p}^{3}+680,pq \right) r+56,{p}^{2}{q}^{ 2}-161,{q}^{3} \ge 0 $ which is concave and quadratic in $r$. – River Li Feb 04 '24 at 09:07
  • @RiverLi In Vietnam, we have some books about pqr. And there are also some Theorem (called MV Theorem or ABC theorem), which said that: if $f(r)$ increases, we need to check inequality for $a=b\ge c$; if $f(r)$ decreases, we need to check for $a=b \le c$ – 30 Anh Ti 711 Feb 04 '24 at 13:46
  • @30AnhTi711 Perhaps someone does not distinguish the pqr and uvw. I saw an article which uses the name pqr, but it actually is the uvw (in my opinion). – River Li Feb 04 '24 at 13:53
  • @RiverLi yes, I also had some confusion at first. Uvw is much further than pqr. – 30 Anh Ti 711 Feb 04 '24 at 14:01
  • @RiverLi Yes, I’ve got the same $pqr$-expression. But now I found out that two of the three inequalities I took from here (https://aritra-12.github.io/pdfs/The_pqr_handout%20(1).pdf) are wrong. $p^3r\geq q^3$ is wrong for sure, and $p^2q^2+12r^2\geq 4pr^3+pqr$ is probably wrong too. Anyway, I was able to finish the proof without referring to $uvw$-theorem or concavity of $f(r)$. Just by using the inequalities on $p,q,r$. – Aig Feb 04 '24 at 20:44
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    @Aig Yes, $p^3r \ge q^3$ is wrong. Some years ago, I found a typo in an inequality book, see my answer here. By the way, it is not difficult without using concave, since $f(r)$ is quadratic. – River Li Feb 05 '24 at 00:52
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    @Aig Now, in my opinion, the pqr method uses explicit relations among $p, q, r$ in the proof, that is, a proof that does not rely on external theorems or lemmas. In particular, sometimes in a pqr proof, we only use well-known inequalities such as $p^2\ge 3q$, $q^2 \ge 3pr$, $p^3 + 9r \ge 4pq$ (perhaps we can call it standard pqr method). The uvw method invokes the uvw theorems in the proof. I think that in the uvw theorems, the well-known inequalities such as $p^2\ge 3q$, $q^2 \ge 3pr$, $p^3 + 9r \ge 4pq$ are not used. Based on these, I distinguish the pqr method from the uvw method. – River Li Feb 05 '24 at 01:31

3 Answers3

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A couple of solutions involving $pqr$-techniques.

Let $p=a+b+c,\;q=ab+bc+ca,\; r=abc.$ Then, after some expanding and simplifying, the inequality turns into:

$$f(r)=-4608r^2+(680pq-63p^3)r+56p^2q^2-161q^3\ge0.$$

We see that $f(r)$ is quadratic and concave. Hence it is sufficient to prove $f(r)\ge 0$ for the minimal and maximal values of $r$. So we need to prove the inequality for $r=0$ and $a=b$.

If $r=0$ then the inequality becomes $56p^2q^2-161q^3\ge0.$ Or $56p^2\ge 161q$, which is true since $p^2\ge 3q.$

If $a=b$ we can put $a=b=1$ since the inequality is homogeneous. The inequality then becomes: $$\frac{2}{7+c}+\frac{1}{7c^2+1}\ge \frac{9}{16c+8}.$$ Expanding and simplifying:

$$224c^3+128c^2+152c+72\ge 63c^3+441c^2+9c+63$$

$$161c^3-313c^2+143c+9 \ge 0$$

The LHS has a root $c=1$. It actually turns out to be of multiplicity $2$. And the LHS can be factorized as $(c-1)^2(161c+9)$ which is non-negative.

Let us prove the initial $pqr$-inequality in a different way: $$680pqr+56p^2q^2\ge 4608r^2+ 63p^3 + 161q^3 .$$

Multiply it by $4$: $$2720pqr+224p^2q^2\ge 18432r^2+ 252p^3 r+ 644q^3 . \quad (*)$$

Consider the following three inequalities:

$$1292pqr+63p^2q^2\ge 16731r^2,\quad \quad \quad(\text{A})$$

$$1134pqr+63p^2q^2\ge 1701r^2+252p^3r+ 252q^3,\quad (\text{B})$$

$$294pqr+98p^2q^2\ge 392q^3.\quad \quad \quad(\text{C})$$

It turns out that inequality $(*)$ is the sum of inequalities (A), (B), (C).

Inequality (A) follows from $pq\ge 9r$ which is the product of two AM-GMs: $p\ge3r^{1/3},\;q\ge3r^{2/3}.$ Then $ 1292pqr+63p^2q^2\ge $ $1292\cdot9r^2+63\cdot9^2\cdot r^2=$ $16731r^2.$

Inequality (B) is just $T(p,q,r)\ge 0$, or $18pqr+p^2q^2\ge 27r^2+4p^3r+ 4q^3$, multiplied by $63$.

Inequality (C) is $p^2q+3pr\ge4q^2$ multiplied by $98q$. The inequality $p^2q+3pr\ge4q^2$ after expanding becomes $\sum a^3b+ab^3\ge \sum2a^2b^2$ which is true due to Muirhead’s inequality.

Aig
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A purely algebraic proof.

WLOG, assume $a \ge b \ge c$. We rewrite the original inequality as: \begin{align*} \left(\frac{1}{7a^2+bc}+\frac{1}{7b^2+ca} - \frac{4}{14ab+bc+ca} \right) + \left( \frac{4}{14ab+bc+ca} + \frac{1}{7c^2+ab}\right) \ge \frac{9}{8\left(ab+bc+ca \right)}.\end{align*} Because: \begin{align*} \frac{1}{7a^2+bc}+\frac{1}{7b^2+ca} - \frac{4}{14ab+bc+ca} = \frac{\left( a-b\right)^2 \left(98ab-21ac-21bc+c^2 \right)}{\left(7a^2+bc \right)\left(7b^2+ca \right) \left(14ab+bc+ca \right)} \ge 0,\end{align*} we only need to prove: \begin{align*}\frac{4}{14ab+bc+ca} + \frac{1}{7c^2+ab} \ge \frac{9}{8\left(ab+bc+ca \right)}. \end{align*} Indeed, by the AM-GM inequality we have: \begin{align*} 7c^2+ab \le \frac{7c^2\left(a+b\right)^2}{4ab}+ab, \end{align*} it's enough to show that: \begin{align*} \frac{4}{14ab+bc+ca} + \frac{4ab}{7\left(bc+ca\right)^2+4a^2b^2} \ge \frac{9}{8\left(ab+bc+ca\right)},\end{align*} or: \begin{align*} \frac{4}{14+x}+\frac{4}{7x^2+4} \ge \frac{9}{8\left(x+1\right)},\end{align*} where $x = \left(bc+ca\right)/ab>0$.

The last inequality is indeed equivalent with $\left(x-2\right)^2\left(161x+18\right) \ge 0$, which is clearly true. We have Q.E.D.

Baby LE
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Remark: I also give a proof using the (standard) pqr method.

Let $p = a + b + c, q = ab + bc + ca, r = abc$. The desired inequality is equivalently written as $$-4608r^2 + (-63p^3 + 680pq)r + 56p^2q^2 - 161q^3 \ge 0. \tag{1}$$

Using $q^2 \ge 3pr$, it suffices to prove that $$-4608r\cdot \frac{q^2}{3p} + (-63p^3 + 680pq)r + 56p^2q^2 - 161q^3 \ge 0$$ or $$q^2(56p^2 - 161q) \ge \left(\frac{1536q^2}{p} + 63p^3 - 680pq\right) r. \tag{2}$$

Since $p^2 \ge 3q$, we have $\mathrm{LHS}_{(2)} \ge 0$. We only need to prove the case that $\frac{1536q^2}{p} + 63p^3 - 680pq > 0$. Using $q^2 \ge 3pr$, it suffices to prove that $$q^2(56p^2 - 161q) \ge \left(\frac{1536q^2}{p} + 63p^3 - 680pq\right) \cdot \frac{q^2}{3p}$$ or $$\frac{q^2(105p^2 + 512q)(p^2 - 3q)}{3p^2} \ge 0$$ which is true using $p^2 \ge 3q$.

We are done.

River Li
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