show this inequality $ab+bc+ac+\sin{(a-1)}+\sin{(b-1)}+\sin{(c-1)}\ge 3$

let $a,b,c>0$ and such $a+b+c=3abc$, show that $$ab+bc+ac+\sin{(a-1)}+\sin{(b-1)}+\sin{(c-1)}\ge 3$$ Proposed by wang yong xi

since $$\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}=3$$ so use Cauchy-Schwarz inequality we have $$(ab+bc+ac)\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}\right)\ge 9$$ so we have $$ab+bc+ac\ge 3$$

Solution 1:

Alright, I think I've finally got a solution, but be warned, it is not a particularly nice one. First off, let's establish the minimum within $0<a,b,c\leq2$, (namely $a=b=c=1$) by using lagrange multipliers, then we'll worry about the case where the numbers can be greater than $2$. So, let's call our constraint function $$g(a,b,c)=a+b+c-3abc$$ $$\nabla g(a,b,c) = [1-3bc,1-3ca,1-3ab]$$ And our function to be minimized $$f(a,b,c)=ab+bc+ca+\sin(a-1)+\sin(b-1)+\sin(c-1)$$ $$\nabla f(a,b,c) = [b+c+\cos(a-1),c+a+\cos(b-1),a+b+\cos(c-1)]$$ Now using lagrange multipliers we can write $$\lambda (1-3bc)=b+c+\cos(a-1)$$ and so on, but to make things easier and symmetrical, let's make a new variable $k=a+b+c=3abc\geq 3$ so we can write instead $$\lambda =h(a)=\frac{k-a+\cos(a-1)}{1-\frac{k}{a}}=-a+\frac{\cos(a-1)}{1-\frac{k}{a}}$$ Note that we can write $\lambda$ in this way replacing $a$ with $b$ or $c$ as it is nice and symmetrical. In order to show that there is one unique solution, I came up with the following argument (which is the only vaguely interesting part of this entire proof, so stop reading after this if you don't want to be bothered with the rest of this mess). If $h(a)$ is a strictly decreasing (or increasing) function over our interval $0<a\leq2$, then in order for $\lambda$ to be constant, we must have $a=b=c$, and if $a=b=c$, then we have $a+a+a=3aaa$ with the only positive solution being $a=b=c=1$ giving $f(a,b,c)=3$ as our local minimum.

MESSY PART Now to show that $h(x)$ is in fact strictly decreasing over our interval $(0,2]$, so let's take the derivative! We want $$0>h'(x)=-1+\frac{-(1-\frac{k}{a})\sin(a-1)-\cos(a-1)\frac{k}{a^2}}{(1-\frac{k}{a})^2}$$ This inequality is implied by $$0>(ka-a^2)\sin(a-1) - k\cos(a-1) - (k-a)^2$$ which is in turn implied by $$0>(k-a)a\sin(1)-k\cos(1) - (k-a)^2$$ since $\sin(1)$ is the maximum of $\sin(a-1)$ and $\cos(1)$ is the maximum of $-\cos(a-1)$ over the interval $a \in (0,2]$ Now this expression is strictly increasing over the interval since $$0<(k-2a)\sin(1)-2a+2k$$ because $$0<-1-4+6 < (3-4)\sin(1)-4+6$$

So our maximum lies at $a=2$, and when we check that against our previous inequality, we get $$0>(2k-4)\sin(1)-k\cos(1) - (k-2)^2$$ And this function is strictly decreasing for $k>3$ since $$0\geq 2+4-2 \cdot 3 - 0 > 2\sin(1)+4-2k - \cos(1)$$ and if we plug in $k=3$ we get a true inequality, that is $$0>(2 \cdot 3-4)\sin(1)-3\cos(1)- (3-2)^2$$ This can be easily proved by bounding $\sin(1)$ and $\cos(1)$ with a short Taylor expansion, namely $\sin(1)=1$ (ending on a positive for the upper bound) and $\cos(1)=1-\frac{1}{2}$ (ending on a negative for the lower bound). So we get $$0>2 - 3 \frac{1}{2} - 1$$ Thus our original inequality holds and the unique extremum is $a=b=c=1$ for $0<a,b,c<2$

And now we must check the boundary cases. Note that this function and constraint is completely symmetrical so we only need to check where $c=2$. Rewriting the inequality we now must prove $$ab+2(a+b)+\sin(a-1)+\sin(b-1)+\sin(1)>3$$ with the constraints $a+b+2=6ab$ and $a,b \leq 2$

We can rewrite the first constraint as $b = \frac{a+2}{6a-1} = 1/6 + \frac{2+1/6}{6a-1}$ and we now we can refine the constraint on $a$ and $b$ since we know that this function of $a$ is decreasing for $a>1/6$ so we know the lowest value of $b$ (and thus $a$) is when $a=2$, so we also have $a,b \geq \frac{4}{11}$ Now the minimum of $ab+2(a+b)=\frac{1}{3}+\frac{13}{6}(a+\frac{1}{6} + \frac{13/6}{6a-1})$ and thus really just $a+b$ can be found by taking the derivative yet again. Doing so we get $$D_a(a+\frac{1}{3}+\frac{13/6}{6a-1})=1-\frac{13}{(6a-1)^2}$$ And setting it equal to $0$ we get the solution $a=b=\frac{1+\sqrt{13}}{6}$. Now for $a>\frac{1+\sqrt{13}}{6}$, since $6a-1$ is clearly increasing, $D_a$ will be postive, and since this is symmetric, this means that $a=b=\frac{1+\sqrt{13}}{6}$ must be the minimum.

Now that we have this, let's plug it in! Doing so we get our minimum to be $$(2+\frac{1}{6})(2\frac{1+\sqrt{13}}{6})+\frac{1}{3}>\frac{13}{6} \cdot \frac{1+3.6}{3}+\frac{1}{3}=\frac{13 \cdot 23 + 30}{90} = \frac{329}{90} = 3 + \frac{59}{90} $$ And now we take our bounds of $a$ to take on the worst-case-scenario for the sinusoidal part of the expression $$2\sin(\frac{4}{11}-1)+\sin(1) > -\frac{10}{11}+1-\frac{1}{6} > -\frac{12-11-2}{12}=-\frac{1}{12}$$ And now we have proved the inequality around the boundary regions, since $\frac{59}{90}>\frac{1}{12}$

Now that that's done, in a similar vein we can show this for our first interval $2<c \leq 3$. Firstly, we assume WLOG that $c \geq a,b$ and now we can establish new bounds for $a$ and $b$ by our constraint $b = \frac{a+c}{3ca-1}$ to get $\frac{2c}{3c^2-1} \leq a,b \leq c$. Note that this minimum is strictly decreasing for $c>2$ since the denominator increases much faster than the numerator, so the lowest point we find is at $c=3$, and that is $\frac{3}{13}$. Now let's again find the minimum of our expression $a+b$ using the derivative again since $$D_a \left(a+\frac{1}{3}+\frac{c+\frac{1}{3c}}{3ca-1}\right)=1-\frac{3c^2+1}{(3ca-1)^2}$$ And setting it equal to $0$ again we get the solution $a=b=\frac{1+\sqrt{1+3c^2}}{3c}$. This is again the minimum since because $3ca-1$ is increasing, $D_a$ will be positive for $a>\frac{1+\sqrt{1+3c^2}}{3c}$, and the function is still symmetric. The minimum of $ab+c(a+b)=(c+\frac{1}{3c})(a+b)$ is thus $$s(c) = \left(c+\frac{1}{3c}\right)\left(2\frac{\frac{1}{c}+\sqrt{\frac{1}{c^2}+3}}{3}\right)+\frac{1}{3}$$ Note that this $s(x)$ is increasing for $c>2$ since $$D_c (s(c)) = \frac{2}{3}\left(1-\frac{1}{3c^2}\right)\left(\frac{1}{c} + \frac{2}{3}\sqrt{\frac{1}{c^2}+3}\right) - \left(c+\frac{1}{3c}\right)\left(\frac{1}{c^2}+\frac{1}{c^3} \frac{1}{\sqrt{\frac{1}{c^2}+3}}\right) > 0 $$ $$\iff \left(c^2-\frac{1}{3}\right)\left(1+\sqrt{1+3c^2}\right)-\left(c^2+\frac{1}{3}\right)\left(1+\frac{1}{\sqrt{1+3c^2}}\right)>0$$ $$\iff -\frac{2}{3}+c^2\left(\sqrt{1+3c^2}-\frac{1}{\sqrt{1+3c^2}}\right)-\frac{1}{3}\left(\sqrt{1+3c^2}+\frac{1}{\sqrt{1+3c^2}}\right) > 0$$ And since $c>2$, this is implied by $$-\frac{2}{3}+c^2\left(\sqrt{1+3c^2}-\frac{1}{3}\right)-\frac{1}{3}\left(\sqrt{1+3c^2}+\frac{1}{3}\right) > 0$$ $$\iff -\frac{8}{9}+\left(c^2-\frac{1}{3}\right)\left(\sqrt{1+3c^2}-\frac{1}{3}\right) > 0$$ And now, since we finally have two increasing functions multiplied together, we can replace each with their lowest value to prove the inequality. Thus since $$\left(4-\frac{1}{3}\right)\left(\sqrt{13}-\frac{1}{3}\right)-\frac{8}{9}>0$$ $s(c)$ is increasing.

Now we have shown that $s(c)$ is increasing and that the minimum of $a,b$ and thus the minimum of $\sin(a-1)$ since $0<a<3$. To prove this inequality for $2<c<3$, all that is left is to do is plug in our worst case scenario. So we have to show $$s(2)-2\sin(\frac{3}{13}-1)+\sin(1)>3$$ Since $\sin(1)$ is the minimum of $\sin(c-1)$ over $2<c \leq 3$. We already calculated $s(2)$ so now we use a few taylor expansions to obtain that this inequality is implied by $$3+\frac{59}{90}-2\left(\frac{10}{13}-\frac{\left(\frac{10}{13}\right)^3}{3!}+\frac{\left(\frac{10}{13}\right)^5}{5!}\right)+1-\frac{1}{6} > 3$$ Which you can do out yourself if you'd like. At this point I will stop doing out these calculations as they make the answer tedious and long.

For our next interval $3<c<1 + \pi$ we can use a better bound for the minimum and maximum of $a,b$ since if $ab+c(a+b)>5$ then it follows that $$ab+c(a+b) + \sin(a-1)+ \sin(b-1) + \sin(c-1) > 5 - 1 - 1 + 0 = 3$$. So all cases where $ab+c(a+b)>5$ are accounted for and this imposing at least the maximum $a,b<a+b<\frac{5}{c}$, and this implies the minimum $a,b>\frac{2(5/c)}{3c(5/c)-1}=\frac{5}{7c}$. Note that this function is still a decreasing one and now we have over the interval $3<c<1+\pi<\frac{7+22}{7}$ that $a,b > \frac{5}{29}$. Using this, we can now prove our inequality by showing $$3 < ab+c(a+b) + \sin(a-1) + \sin(b-1) + \sin(c-1) < s(3) + 2\sin(5/29-1) < s(3) -48/29 \approx 3.02784 $$ And this can be trivially shown by bounding $s(3)$ as well.

Next interval: $1+\pi < c < 1 + \pi + 1$. We can no longer use the bound from the previous problem, but we can at this point just say $$s(1+\pi) + \sin(a-1)+\sin(b-1)+\sin(c-1) > s(4) -3\sin(1) \approx 3.25 > 3$$ This can again easily be done using a quick taylor expansion, and that $s(4)=\frac{52}{9}$

Finally for our interval $c>2+\pi$ we finally have $s(2+\pi)>6$, so we are done.

Note: There is probably at least one mistake in here, given the length, so feel free to leave comments.

Solution 2:

$$\mathbf{\color{brown}{Analysis\ of\ the\ issue\ condition.}}$$

Let WLOG $$0<a\le b\le c,\tag1$$ $$a+b+c=3abc,\tag2$$ then $$c+a=(3ac-1)b,\quad (3ac-1)a\le a+c\le(3ac-1)c,$$ $$\begin{cases} (3a^2-1)c\le2a\\[4pt] (3c^2-1)a\le2c\\[4pt] 0<a\le b\le c, \end{cases}\rightarrow \begin{cases} \dfrac1{\sqrt3} < a\le \dfrac{2c}{3c^2-1}\\ a\le c\le\dfrac{2a}{3a^2-1}\\ \dfrac{2c}{3c^2-1}\le a\le c\\ \end{cases}\rightarrow \begin{cases} \dfrac1{\sqrt3}< a\\[4pt] (\sqrt3c-1)^2\le2\\[4pt] 3a^2-1\le 2\le 3c^2-1\\[4pt] \end{cases}$$ $$\boxed {\dfrac1{\sqrt3}<a\le1\le c\le\dfrac{\sqrt2+1}{\sqrt3},\quad b=\dfrac{a+c}{3ac-1}.}\tag3$$ Besides, as shown in OP, $$\dfrac13\left(\dfrac1{ab}+\dfrac1{bc}+\dfrac1{ca}\right)=1.$$ Using AM-HM, easy to obtain $$ab+bc+ac\ge3.\tag4$$

$$\mathbf{\color{brown}{The\ trig\ inequality}}$$

Let us consider the inequality $$\sin(a-1) + \sin(b-1) + \sin(c-1) +\lambda(a+b+c -3abc)\tag5$$ under the constraints $(3).$

Using Lagrange multipiers method with the function $$f(a,b,c,\lambda) = \sin(a-1) + \sin(b-1) + \sin(c-1) +\lambda(a+b+c -3abc),$$ the stationary points can be found from the system $f'_a=f'_b=f'_c=f'_\lambda=0,$ or \begin{cases} \cos(a-1)+ \lambda(1-3bc)=0\\ \cos(b-1)+ \lambda(1-3ca)=0\\ \cos(c-1)+ \lambda(1-3ab)=0\\ a+b+c=3abc\\ \dfrac{\sqrt3}3<a\le1\le c\le\dfrac{\sqrt2+1}{\sqrt3},\\ \end{cases} \begin{cases} \dfrac1{3bc-1}\cos(a-1) - \lambda = 0\\ \dfrac1{3ca-1}\cos(b-1) - \lambda = 0\\ \dfrac1{3ab-1}\cos(c-1) - \lambda = 0\\ a+b+c=3abc\\ \dfrac{\sqrt3}3<a\le1\le c\le\dfrac{\sqrt2+1}{\sqrt3},\\ \end{cases} \begin{cases} b =\dfrac{a+c}{3ac-1}\\[4pt] 3bc-1=\dfrac{3ac+3c^2}{3ac-1}-1 = \dfrac{3c^2+1}{3ac-1}\\[4pt] 3ab-1=\dfrac{3a^2+3ac}{3ac-1}-1 = \dfrac{3a^2+1}{3ac-1}\\[4pt] \dfrac{3ac-1}{3c^2+1}\cos(a-1) = \dfrac{3ac-1}{3a^2+1}\cos(c-1)\\[4pt] \dfrac{3ac-1}{3c^2+1}\cos(a-1) = \dfrac1{3ac-1}\cos\left(\dfrac{a+c}{3ac-1}-1\right)\\[4pt] \dfrac{\sqrt3}3<a\le1\le c\le\dfrac{\sqrt2+1}{\sqrt3}.\tag6 \end{cases} Function $g(x)=(3x^2+1)\cos (x-1)$ has positive derivation $$g'(x)=6x\cos(x-1)-(3x^2+1)\sin(x-1)$$ in $\left[\dfrac1{\sqrt3},\dfrac{\sqrt2+1}{\sqrt3}\right],$ so $g(x)$ increases in this interval.

The equation $(6.4)$ can be rewrited as $$g(a)=g(c).$$ Taking in account constraints $(6.6),$ the system $(6)$ can be satisfied in the single point $$a=1,\quad c=g(1)=1.$$ This gives the single stationary point $$a=b=c=1,\quad f(a,b,c,\lambda)=0.$$

Note that constraints $(6.6)$ determines the blurring of boundaries between variables and can not be checked. On the other hand, checking of the trig inequality in the random point $$\left(a,b,c, f\right)=\left(\dfrac23,\dfrac65,\dfrac43, \sin\dfrac15\right)$$ shows that trig inequality holds. Therefore, it is proved.

Taking in account $(4)$, the issue inequality is proved.