Nice, interesting and beautiful. I will prove the stronger inequality: $(a^{2}+2)(b^{2}+2)(c^{2}+2) \geq 3(a+b+c)^{2}.$ I will use the fact that if x,y have the same sign, then (1+x)(1+y)>=1+x+y. Write the inequality as $\prod \left(\frac{a^{2}-1}3+1\right) \geq \frac{1}{9}(a+b+c)^{2}.$ If a,b,c>=1 then $\left(\frac{a^{2}-1}3+1\right) \geq 1+\sum \frac{a^{2}-1}3= \frac{1}{3}\sum a^{2}\geq \frac{1}{9}(a+b+c)^{2}.$ If two of them are at least 1, let them be a and b. Then the LHS is at least $( 1+\frac{1}{3}(a^{2}-1+b^{2}-1) ) (c^{2}+2)= \frac{1}{9}(a^{2}+b^{2}+1) (1^{2}+1^{2}+c^{2}) \geq (a+b+c)^{2}$ by Cauchy. If one of them is at least 1, let it be a, then we use the same argument as in the precedent case. If a,b,c<=1 then by Bernoully inequality we have $\prod \left(1+\frac{a^{2}-1}3\right) \geq 1+\sum \frac{a^{2}-1}3= \frac{1}{3}\sum a^{2}\geq \frac{1}{9}(a+b+c)^{2}.$

There is a simple lower-power solution using strictly AM-GM: productcyc(a^2+2) =((abc)^2+1+1) + sumcyc(a^2) + 3sumcyc(a^2) + 2sumcyc((ab)^2+1)) >=3(abc)^2/3 + sumcyc(a^2) + 3sumcyc(ab) + 4sumcyc(ab) >=sumsym(a^4/3*b^2/3) + 7sumcyc(ab) (expanding the well known inequality xyz>=productcyc(x+y-z), where x=a^2/3, y=b^2/3, x=c^2/3) >=2sumcyc(ab)+7sumcyc(ab) =9sumcyc(ab), proving the required. Equality obviously holds iff a=b=c=1. Of course, even though it's simpler, Harazi's result is more elegant in another sense because it is stronger.

Particular case of http://www.mathlinks.ro/Forum/viewtopic.php?t=19666 for k = 1. darij

We have to prove that $\frac{ab+bc+ca}{(a^{2}+2)(b^{2}+2)(c^{2}+2)}\le \frac{1}{9}$ Now $x=\frac{1}{a^{2}+2},...$ Use AM-GM $2\frac{1}{(c^{2}+2)}\frac{a}{(a^{2}+2)}\frac{b}{(b^{2}+2)}\le\frac{1}{c^{2}+2}(\frac{a^{2}}{(a^{2}+2)^{2}}+\frac{b^{2}}{(b^{2}+2)^{2}})= z(x-2x^{2}+y-2y^{2})...$ And we prove $xy+yz+zx\le x^{2}y+y^{2}x+y^{2}z+z^{2}y+z^{2}x+x^{2}z+\frac{1}{9}$ But this is easy: By AM-GM we have $x^{2}y+xy^{2}+\frac{1}{27}\geq xy$, ...

Does anyone have the *OFFICIAL* solutions to this question on the APMO? I remembere there were 2 nice solutions, one with vectors, one with trig.

@Michael Lipnowski :please edit it i don't understand

mathboy_cntHDVN wrote: @Michael Lipnowski :please edit it i don't understand $\prod_{cyc}(a^{2}+2) =((abc)^{2}+1+1)+\sum_{cyc}(a^{2})+3 \sum_{cyc}(a^{2})+2 \sum_{cyc}((ab)^{2}+1))$ $\geq 3(abc)^\frac{2}{3}+\sum_{cyc}(a^{2})+3 \sum_{cyc}(ab)+4 \sum_{cyc}(ab) \geq \sum_{sym}(a^\frac{4}{3}{b^\frac{2}{3})+7 \sum_{cyc}(ab)}$ (expanding the well known inequality $xyz\geq \prod_{cyc}(x+y-z)$, where $x=a^\frac{2}{3}, y=b^\frac{2}{3}, x=c^\frac{2}{3})$ $\geq 2 \sum_{cyc}(ab)+7 \sum_{cyc}(ab) =9 \sum_{cyc}(ab)$, proving the required. Equality obviously holds iff $a=b=c=1$. @mathboy_cntHDVN : This 's posts of Michael Lipnowski in latex

My solution: Put $ t = \frac {b + c}{2}$ By AM-GM we have: $ (b^2 + 2)(c^2 + 2) \geq 3 + 2bc + 2(b^2 + c^2) \geq 3 + 6t^2$ and $ t^2 \geq bc$ We need prove that: $ (a^2 + 2)(3 + 6t^2) \geq 9(2at + t^2)$ <=>$ (a - t)^2 + 2(at - 1)^2 \geq 0$ This inequality is true Equality occurs when $ a = b = c = 1$

Arne wrote: Prove that the inequality $ \left(a^{2} + 2\right)\left(b^{2} + 2\right)\left(c^{2} + 2\right) \geq 9\left(ab + bc + ca\right)$ holds for all positive reals $ a$, $ b$, $ c$. I will prove the stronger inequality: $ (a^2 + 2)(b^2 + 2)(c^2 + 2) \ge 3(a + b + c)^2$ I have: $ (a^2 + 2)(b^2 + 2) = (a^2 + 1)(b^2 + 1) + a^2 + b^2 + 3 \ge (a + b)^2 + \frac {1}{2}(a + b)^2 + 3$ $ = \frac {3}{2}[(a + b)^2 + 2]$ $ = > (a^2 + 2)(b^2 + 2)(c^2 + 2) \ge \frac {3}{2}[(a + b)^2 + 2](c^2 + 2) \ge \frac {3}{2}[ \sqrt {2}(a + b) + \sqrt {2}c]^2$ $ = 3(a + b + c)^2$ $ =>(a^2 + 2)(b^2 + 2)(c^2 + 2) \ge 3(a + b + c)^2 \ge 9(ab + bc + ca)$

Can anyone please explain in more detail how Bernoulli's inequality is used in Harazi's solution? Thank you in advance!

there was sdrfgdf

We have $(\forall {a,b,c\in\mathbb{R}})(a^2+2)(b^2+2)(c^2+2)\ge{4(a^2+b^2+c^2)+5(ab+bc+ca)}.$ See: http://www.irgoc.org/viewtopic.php?f=10&t=926&p=8055&sid=8b67e9ff2c1d570b87e6702457829c68#p8055 http://www.irgoc.org/viewtopic.php?f=10&t=926&sid=5d489375ae9166d3d2fb7b5f3fe0763a&start=220

Take $ x=a+b+c $ , $ y=ab+bc+ca $ , $ z=abc $. Then given inequality is equivalent to \[ z^{2}+2y^{2}-4xz+4x^{2}-17y+8\geq 0 \] It is equivalent to \[ (z-\frac{x}{3})^{2}+\frac{8}{9}(y-3)^{2}+\frac{10}{9}(y^{2}-3xz)+\frac{35}{9}(x^{2}-3y)\geq 0 \] It is easy to prove \[ y^{2}-3xz\geq 0 \] and \[ x^{2}-3y\geq 0 \] I took it from official solutions. There were another two solutions. One of them is by Jensen , other is by vectoral product of vectors.

mudok wrote: Take $ x=a+b+c $ , $ y=ab+bc+ca $ , $ z=abc $. Then given inequality is equivalent to \[ z^{2}+2y^{2}-4xz+4x^{2}-17y+8\geq 0 \] It is equivalent to \[ (z-\frac{x}{3})^{2}+\frac{8}{9}(y-3)^{2}+\frac{10}{9}(y^{2}-3xz)+\frac{35}{9}(x^{2}-3y)\geq 0 \] May anyone explain to me how would one think of grouping the addends like that? I mean, there is no formula nor anything obvious in this particular grouping right? So how can you master the craft of grouping, even if it works only in some cases? And, what "shows" you how to group the elements in this problem? Thank you in advance!

I didn't solve it.I said that I took it from official solutin. That's why I don't know the way of finding this solution. But to get \[ z^{2}+2y^{2}-4xz+4x^{2}-17y+8\geq 0 \] open parantheses the given given inequality and substitute \[ (ab)^{2}+(bc)^{2}+(ca)^{2}=y^{2}-2xz \] \[ a^{2}+b^{2}+c^{2}=x^{2}-2y \] \[ (abc)^{2}=z^{2} \]

mudok wrote: I didn't solve it.I said that I took it from official solutin. That's why I don't know the way of finding this solution. But to get \[ z^{2}+2y^{2}-4xz+4x^{2}-17y+8\geq 0 \] open parantheses the given given inequality and substitute \[ (ab)^{2}+(bc)^{2}+(ca)^{2}=y^{2}-2xz \] \[ a^{2}+b^{2}+c^{2}=x^{2}-2y \] \[ (abc)^{2}=z^{2} \] I understand very well how \[ z^{2}+2y^{2}-4xz+4x^{2}-17y+8\geq 0 \]was obtained, but thank you for the explanation. I also didn't expect you to know how the groupings were made, but just hoped that someone might understand the reasoning behind this solution and be kind enough to explain.

StefanS: I understand very well how \[ z^{2}+2y^{2}-4xz+4x^{2}-17y+8\geq 0 \]was obtained, but thank you for the explanation. I also didn't expect you to know how the groupings were made, but just hoped that someone might understand the reasoning behind this solution and be kind enough to explain.[/quote] To StefanS: 1)Not at all. 2)My english is bad.Sorry.

Arne wrote: Prove that the inequality $\left(a^{2}+2\right)\left(b^{2}+2\right)\left(c^{2}+2\right) \geq 9\left(ab+bc+ca\right)$ holds for all positive reals $a$, $b$, $c$. Inequality stronger is true \[\left(a^{2}+2\right)\left(b^{2}+2\right)\left(c^{2}+2\right) \geq 3\left(a+b+c\right)^2+(abc-1)^2\]

Nguyenhuyen_AG wrote: Inequality stronger is true \[\left(a^{2}+2\right)\left(b^{2}+2\right)\left(c^{2}+2\right) \geq 3\left(a+b+c\right)^2+(abc-1)^2\] Beacause $ \iff \frac{2}{3}(ab+bc+ac-3)^2+\left[ a^2+b^2+c^2+2abc+1-2(ab+bc+ac) \right]+ \left( \frac{4}{3}(ab+bc+ac)^2-4abc(a+b+c) \right) \geq 0$

Arne wrote: Prove that the inequality \[\left(a^{2}+2\right)\left(b^{2}+2\right)\left(c^{2}+2\right) \geq 9\left(ab+bc+ca\right)\]holds for all positive reals $a$, $b$, $c$. WLOG $(a^2-1)(b^2-1)\geq 0, $ then $$(a^2+2)(b^2+2)(c^2+2)\ge3(a^2+b^2+1)(1+1+c^2)\ge3(a+b+c)^2\ge9(ab+bc+ca)$$

sqing wrote: Arne wrote: Prove that the inequality \[\left(a^{2}+2\right)\left(b^{2}+2\right)\left(c^{2}+2\right) \geq 9\left(ab+bc+ca\right)\]holds for all positive reals $a$, $b$, $c$. WLOG $(a^2-1)(b^2-1)\geq 0, $ then $$(a^2+2)(b^2+2)(c^2+2)\ge3(a^2+b^2+1)(1+1+c^2)\ge3(a+b+c)^2\ge9(ab+bc+ca)$$ Very nice way of proving it. Congratulations on finding it!

Let $a, b, c$ be positive reals. Prove that $$\left(a^{2}+2\right)\left(b^{2}+2\right)\left(c^{2}+2\right) \geq 4\left(a^2+bc+ca\right)$$$$\left(a^{2}+2\right)\left(b^{2}+2\right)\left(c^{2}+2\right) \geq 4\left(a^2+bc+c^2\right)$$

$8<9$ We first e x p a n d which yields the equivalent inequality $$a^2b^2c^2+2\sum_{\mathrm{cyc}} a^2a^2+4\sum_{\mathrm{cyc}} a^2-9\sum_{\mathrm{cyc}}ab+8=a^2b^2c^2+2\sum_{\mathrm{cyc}} (ab-1)^2+4\sum_{\mathrm{cyc}} a^2-5\sum_{\mathrm{cyc}} ab+2\geq 0.$$Motivated by the $a=b=c=1$ equality case we discard the $(ab-1)^2$ terms, so it remains to prove that $$4\sum_{\mathrm{cyc}} a^2+a^2b^2c^2+2\geq 5\sum_{\mathrm{cyc}}ab \impliedby \sum_{\mathrm{cyc}} a^2+3a^{\frac{2}{3}}b^{\frac{2}{3}}c^{\frac{2}{3}}\geq 2\sum_{\mathrm{cyc}} ab$$by AM-GM. By Schur on $a^{\frac{2}{3}}$, etc., we have $$\sum_{\mathrm{cyc}} a^2+3a^{\frac{2}{3}}b^{\frac{2}{3}}c^{\frac{2}{3}}\geq \sum_{\mathrm{sym}} a^{\frac{4}{2}}b^{\frac{2}{3}},$$from which the conclusion is clear by AM-GM. $\blacksquare$ Here is an alternate finish from the second-to-last centered inequality. The inequality there is homogeneous and symmetrical, so WLOG suppose that $a\geq b \geq c$ and $ab=1$. We will prove that the inequality is strongest when $a=b=1$. The inequality in this case is equivalent to $$\left(a+\frac{1}{a}\right)^2-2c\left(a+\frac{1}{a}\right)+c^2+3c^{\frac{2}{3}}-4\geq 0.$$Viewing the LHS as a quadratic in $a+\tfrac{1}{a}$, its minimum occurs when $a+\tfrac{1}{a}=c$ and is decreasing on $[c,\infty)$ and thus on $[2,\infty)$ since $a\geq b\geq c$. This proves the desired claim, since $a+\tfrac{1}{a} \implies a=1$. The inequality then reduces to $c^2-4c+3c^{\frac{2}{3}}\geq 0$, which is just AM-GM on $c^2$ and three copies of $c^{\frac{2}{3}}$. $\blacksquare$

Arne wrote: Prove that the inequality \[\left(a^{2}+2\right)\left(b^{2}+2\right)\left(c^{2}+2\right) \geq 9\left(ab+bc+ca\right)\]holds for all positive reals $a$, $b$, $c$. 2023 Shandong High School Mathematics Competition Q13

Solution is attached. Although a bit lengthy, I found this solution to be very natural.

Attachments:

Version 1- 4_Variables Let $a,b,c,d$ be positive reals. Then prove that $$(a^2+2)(b^2+2)(c^2+2)(d^2+2)\geq \dfrac{9(a+b+c+d)^2}{2}\geq 18(a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{4}+a_{4}a_{1})$$

Version 2- 4_Variables Let $a,b,c,d,f$ be positive reals. Then prove that $$(a^2+2)(b^2+2)(c^2+2)(d^2+2)(f^2+2)\geq \dfrac{27(a+b+c+d+f)^2}{4}$$

Gen1 was here

These generalizations are proving the stronger one.

Gen2 was here

Nice. Upon expanding the left side and using the inequalities $a^2b^2+1\ge 2ab,a^2b^2c^2+2\ge3\sqrt[3]{a^2b^2c^2}$ from AM-GM we want to prove $4(a^2+b^2+c^2)+3\sqrt[3]{a^2b^2c^2}-5(ab+bc+ca)\ge 0.$ This follows from \begin{align*}&\frac32\left((a-b)^2+(b-c)^2+(c-a)^2\right)+\left((\sqrt[3]a-\sqrt[3]b)^2\sqrt[3]{a^2b^2}+(\sqrt[3]b-\sqrt[3]c)^2\sqrt[3]{b^2c^2}+(\sqrt[3]c-\sqrt[3]a)^2\sqrt[3]{c^2a^2}\right)\\ &+\sqrt[3]{a^2}(\sqrt[3]{a^2}-\sqrt[3]{b^2})(\sqrt[3]{a^2}-\sqrt[3]{b^2})+\sqrt[3]{b^2}(\sqrt[3]{b^2}-\sqrt[3]{a^2})(\sqrt[3]{b^2}-\sqrt[3]{c^2})+\sqrt[3]{c^2}(\sqrt[3]{c^2}-\sqrt[3]{a^2})(\sqrt[3]{c^2}-\sqrt[3]{b^2})\\ &\ge 0.\end{align*}The last part is by Schur. Equality holds at $a=b=c=1.$

Another solution. Again expand but notice $a^2b^2c^2+\frac13(a^2+b^2+c^2)+2\ge\frac23(a^2bc+b^2ca+c^2ab)+2\ge\frac43(a\sqrt{bc}+b\sqrt{ca}+c\sqrt{ab})$ and combined with $a^2b^2+1\ge 2ab$ we want to show $4(a\sqrt{bc}+b\sqrt{ca}+c\sqrt{ab})+11(a^2+b^2+c^2)-15(ab+bc+ca)\ge 0.$ This follows from \begin{align*}&\frac72\left((a-b)^2+(b-c)^2+(c-a)^2\right)+4(\sqrt{ab}(\sqrt a-\sqrt b)^2+\sqrt{bc}(\sqrt b-\sqrt c)^2+\sqrt{ca}(\sqrt c-\sqrt a)^2)\\&+4\left(a(\sqrt a-\sqrt b)(\sqrt a-\sqrt c)+b(\sqrt b-\sqrt a)(\sqrt b-\sqrt c)+c(\sqrt c-\sqrt a)(\sqrt c-\sqrt b)\right)\\&\ge 0.\end{align*}The last part again by Schur.

Inspired by the equality case $a=b=c=1$, we first notice \[(a^2+2)(b^2+2) = (a^2+1)(b^2+1) + (a^2+b^2) + 3 \ge \frac 32 (a+b)^2+3,\] so our lower bound for the LHS is \begin{align*} \frac 32 \left((a+b)^2+2\right)(c^2+1) &= \frac 32 \left((ac+bc)^2+4+2\left((a+b)^2+c^2\right)\right) \\ &\ge \frac 32 \left(4c(a+b)+2c^2+2(a+b)^2\right) \\ &= 3(a+b+c)^2 \ge 9(ab+bc+ca). \quad \blacksquare \end{align*}

Let $a, b, c\geq 0$ and fix $a+b+c=t$. We claim that it suffices to prove the inequality when two of $a,b,c$ are equal. Define $f,g$ over $[0,t]^3$ by $$f(a,b,c,d)=(a^2+2)(b^2+2)(c^2+2)-9(ab+bc+ca)$$and $g(a,b,c)=a+b+c$. Since the set $\{(a,b,c)\in [0,t]^3:g(a,b,c)=t\}$ is compact, $f$ achieves a minimum value in this set. If that minimum is achieved at the endpoints, then our claim clearly follows. Otherwise suppose the minimum is attained at the points $a,b,c\in (0,t)$ then by Lagrange Multipliers, $$\nabla f=\lambda \nabla g \Rightarrow \begin{cases} 2(a+1)(b^2+2)(c^2+2)-9(b+c)=\lambda \\ 2(b+1)(c^2+2)(a^2+2)-9(c+a)=\lambda \\ 2(c+1)(a^2+2)(b^2+2)-9(a+b)=\lambda \end{cases}.$$Now, suppose $a,b,c$ are pairwise distinct. Subtracting the second equation from the first and dividing both sides by $b-a$ we get $$2(c^2+2)(ab+a+b-2)=9$$and we get two other similar equations too. Let $\alpha=abc, \beta=ab+bc+ca, \gamma=a+b+c$. We can check that $$0=(a^2+2)(bc+b+c-2)-9/2=a(\alpha+\beta-2\gamma-2)-\alpha+2\beta+2\gamma-9/2-4 \qquad (1)$$Since the same holds with $a$ replaced by $b$ in $(1)$, it follows that the coefficient of $a$ in $(1)$ must be zero since $a\neq b$. Thus we get, $$\alpha+\beta-2\gamma-2=0, -\alpha+2\beta+2\gamma-9/2-4=0.$$So $\beta=7/2$ and $\alpha=2\gamma-3/2$. Now, \begin{align*} & (c^2+2)(ab+a+b-2)=\frac 92 \\ \Rightarrow & (c^2+2)(ab+\gamma-c-2)=\frac 92 \\ \Rightarrow & -c^3+c^2(\gamma-2)+c(abc-2)+2\gamma+2ab=4+\frac 92\\ \Rightarrow & -c^3+c^2(\gamma-2)+c(2\gamma-3/2-2)+2\gamma+2ab=4+\frac 92.\end{align*}So by symmetry we get $$-(a^3+b^3+c^3)+(a^2+b^2+c^2)(\gamma-2)+(a+b+c)(2\gamma-3/2-2)+6\gamma+2(ab+bc+ca)=3\left (4+\frac 92\right).$$Putting everything above in terms of $\alpha,\beta,\gamma$ allows us to solve for $\gamma$. We get, if my calculations are right, $\gamma=0$. But this is clearly a contradiction since $a,b,c>0$. And so our claim follows. \ So it remains to prove the inequality when two variables are equal. Indeed note that $$(x^2+2)^2(y^2+2)-9(x^2+2yx)=\frac{1}{(x^2+2)^2} \left ((y(x^2+2)^2-9x)^2+(x^2-1)^2 (2x^4+11x^2+32)\right)\geq 0.$$