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$a,b,c >0$, and $a^ab^bc^c=1$, prove $$a^2b+b^2c+c^2a \leqslant 3$$

I don't even know what to do with the condition $a^ab^bc^c=1$. At first I think $x^x>1$, but I was wrong. This inequality is true, following by the verification from Mathematica

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    $\begingroup$ I haven't really worked on this much, but if you could somehow find$\|(a^2,b^2,c^2)\|$ and $\|(b,c,a)\|$ then you would presumably be done by Cauchy-Schwarz because $\langle(a^2,b^2,c^2),(b,c,a)\rangle = a^2b + b^2c + c^2a$. $\endgroup$ – kcborys May 6 '16 at 23:40
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    $\begingroup$ Does somebody have any idea how Mathematica did that verification? $\endgroup$ – Han de Bruijn Jul 5 '16 at 13:39
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+25
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We employ of the rearrangement inequality.
First, since $x\mapsto x^2$ preserves the order for $x>0,$ we have $$a^2b+b^2c+c^2a\le a^3+b^3+c^3.$$
Next write the condition as $$a\ln a+b\ln b+c\ln c=0.$$ Again using the rearrangement inequality, as $x\mapsto\ln x$ preserves the order for $x>0,$ we have the inequalities: $$\begin{cases}a\ln b+b\ln c+c\ln a\le0\\ a\ln c+b\ln a+c\ln b\le0\\ a\ln a+b\ln b+c\ln c=0\end{cases}$$
Adding these together, we have $(a+b+c)\ln(abc)\le0,$ hence $abc\le1.$
Now let $\ln a+\ln b+\ln c=k\le0.$ Apply the Lagrange multiplier method with condition $g(a,b,c):=\ln a+\ln b+\ln c=k$ for a fixed $k\le0,$ to maximize $f(a,b,c):=a^3+b^3+c^3.$
Thus the Lagrange multiplier gives that the extreme of $f$ occurs when $$\begin{cases}3a^2-\lambda\frac{1}{a}=0\\ 3b^2-\lambda\frac{1}{b}=0\\ 3c^2-\lambda\frac{1}{c}=0\end{cases},$$ i.e. when $a^3=b^3=c^3=\lambda;$ then $a=b=c$ and hence $f(a,b,c)=3abc=3e^k\le3.$
$\square$

Hope this helps.

Edit:
As pointed out in the comment, the final part about Lagrange multipliers is incorrect; in fact, given the constraint $abc<1,$ it does not follow that $a^2b+b^2c+c^2a\le3.$
Also pointed out in the comment is to use Jensen's inequality to obtain $a+b+c\le3,$ but we are not getting answers yet.
Everything we tried so far to fix this fails. We shall update if we find a way to work around.

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    $\begingroup$ Lagrange multiplier doesn't prove that we obtain a global maximum. Note that we need to check the boundaries, which is no easy case at all. Actually see that $a=5, b=\frac 15, c = e^{k}$ satisfy the constraint, while $f(a,b,c) = 125 + \frac{1}{125} + e^{3k} > 3$ $\endgroup$ – Stefan4024 Jul 4 '16 at 22:18
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    $\begingroup$ Actually in fact using $\ln(abc) = k$ for non-positive values of $k$ isn't a good constrain, as it relaxes the inital one and it makes the inequality false under that assumption. $\endgroup$ – Stefan4024 Jul 4 '16 at 22:20
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    $\begingroup$ @awllower I don't think that there's something's wrong with both friends having MSE accounts, as long as they are not intentionally voting on each other's answers in order to increase reputation. On the other side taking logarithm and applying Jensen's Inequality on $x ln(x)$ we get that $a+b+c \le 3$. Maybe you can use this and try to put two constraints on Lagrange's Multiplier. $\endgroup$ – Stefan4024 Jul 5 '16 at 5:42
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    $\begingroup$ @Stefan4024 I am afraid these two constraints don't suffice: $$2.46375+\frac{1}{2}+\varepsilon<3\\2.46375\cdot\frac{1}{2}\cdot\varepsilon<1\\(2.46375)^2/2\cong3.035>3.$$ But thanks for this idea! $\endgroup$ – awllower Jul 5 '16 at 6:04
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    $\begingroup$ @yoyo and awllower, the inequality does not hold as soon as you switch to $a^3+b^3+c^3$, e.g., take any $0.25\le a\le 0.5$ and $b=c$. From $a^ab^bc^c=1$, we have $b=c>1.16$, but $1.16^3+1.16^3>3$, so I am not sure if this approach is fixable. $\endgroup$ – Wiley Jul 16 '16 at 18:34
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We shall prove the following inequalities:

If $x+y+z=3$ and $x,y,z>0$, then $$\tag{1}x^2y+y^2z+z^2x+xyz\le 4\label{1},$$ $$\tag{2}3x^xy^yz^z+xyz\ge 4\label{2},$$ i.e., $$\tag{*}x^2y+y^2z+z^2x\le 3x^xy^yz^z\label{*}.$$

Accept $\eqref{*}$ for a moment. Homogenizing the inequality by substitution

$$x=\frac{3a}{a+b+c},~y=\frac{3b}{a+b+c},~z=\frac{3c}{a+b+c},$$

we have $\forall a,b,c>0$,

$$\tag{**}a^2b+b^2c+c^2a\le 3(a^ab^bc^c)^{3/(a+b+c)}.\label{**}$$

If $a^ab^bc^c=1$, then we get the original inequality as desired.


Proof of $\eqref{1}$:

Note the LHS is cyclic, we cannot assume a specific order like $x\le y\le z$, but we can assume WLOG that $y$ is in the middle (neither the minimum nor the maximum), then

$$z(y-x)(y-z)\le 0\\ \Rightarrow x^2y+y^2z+z^2x+xyz\le y(x^2+2xz+z^2)=\frac{1}{2}(2y)(x+z)^2\le\frac{1}{2}\left(\frac{2(x+y+z)}{3}\right)^3=4,$$ by AM-GM. Equality holds when $(x,y,z)=(1,1,1),(2,1,0)$ along with the cyclic permutations.


Proof of $\eqref{2}$:

WLOG we assume $x=\min(x,y,z)$ and consider the following two cases:

  • $x\ge 1/3$, then $y,z\ge 1/3$. We note that the function

$$f(t)=\left(t+\frac{1}{3}\right)\ln t$$

is convex for $t\ge 1/3$, as

$$f''(t)=\frac{3t-1}{3t^2}\ge 0.$$

By Jensen,

$$ \begin{align} &~~~~~~~x^{x+1/3}y^{y+1/3}z^{z+1/3}=\exp\left\{f(x)+f(y)+f(z)\right\} \ge\exp\left\{3f\left(\frac{x+y+z}{3}\right)\right\}=\exp\left\{3f(1)\right\}=1,\\ &\Rightarrow x^xy^yz^z\ge (xyz)^{-1/3},\\ &\Rightarrow 3x^xy^yz^z+xyz\ge 3(xyz)^{-1/3}+xyz\ge 4\sqrt[4]{(xyz)^{-1}(xyz)}=4, \end{align} $$

where we have applied AM-GM for the last line. Equality holds when $x=y=z=1$.

  • $0<x\le1/3$. We note that the function

$$g(t)=t\ln t$$

is convex for $t>0$, as

$$g''(t)=\frac{1}{t}> 0.$$

Again by Jensen, we have

$$ \begin{align} &~~~~~~~y^yz^z=\exp\left\{g(y)+g(z)\right\}\ge\exp\left\{2g\left(\frac{y+z}{2}\right)\right\} =\exp\left\{2g\left(\frac{3-x}{2}\right)\right\},\\ &\Rightarrow x^xy^yz^z\ge\exp\left\{g(x)+2g\left(\frac{3-x}{2}\right)\right\} \ge\exp\left\{g\left(\frac{1}{3}\right)+2g\left(\frac{4}{3}\right)\right\} =\frac{32}{27}\sqrt[3]{2}>\frac{4}{3},\\ &\Rightarrow 3x^xy^yz^z+xyz>4+xyz>4. \end{align} $$

The second line follows from that $h(t):=g(t)+2g((3-t)/2)$ is monotonically decreasing for $t\in(0,1)$, so $h(x)\ge h(1/3)$ as $x\le 1/3$. Proof of monotonicity:

$$h'(t)=\ln\left(\frac{2t}{3-t}\right)<0\iff 0<t<1.$$


Post-mortem:

  1. Despite the flow of the proof, the starting point is the homogeneous version \eqref{**};

  2. \eqref{1} is due to Vasile, which allows us to strengthen the inequality into a symmetric one as \eqref{2} (I am not able to find the originial post on artofproblemsolving, but see e.g., here). It seems that converting \eqref{*} into other symmetric form such as $x^3+y^3+z^3$ is too strong for it to hold;

  3. For \eqref{2}, the case of $x\le 1/3$ is unfortunately necessary, as $x^xy^yz^z\ge(xyz)^{-1/3}$ does NOT always hold, e.g., check $x\to0,y-z\to 0$. Is there any better estimate of $x^xy^yz^z$ so we can avoid the derivatives? Another follow-up question is to determine the smallest $k$ such that $kx^xy^yz^z+xyz\ge k+1$ holds given the constraints. It definitely fails for $k\le 1/2$.

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Brute Force ($200\times200\times200$ grid) Not a proof, but couldn't resist.
Of course, everybody knows where the maximum is, for reasons of symmetry:
with $\,(a,b,c) = (1,1,1)\,$ we have $\,a^ab^bc^c=1\,$ and $\,a^2b+b^2c+c^2a = 3$ .
The following (Delphi Pascal) program is supposed to be self documenting.

program HN_NH;
{
  Brute Force with Seven Point Stars
  ==================================
}
function pow(x,r : double) : double;
{
  x^r
}
begin
  pow := exp(r*ln(x));
end;
procedure test(veel : integer); var i,j,k,ken : integer; a,b,c,d,f,min,max : double;
procedure vertex(x,y,z : integer); var a,b,c,h : double; begin a := (2*i+x)*d; b := (2*j+y)*d; c := (2*k+z)*d; h := pow(a,a)*pow(b,b)*pow(c,c); if h < 1 then ken := ken*2 else ken := ken*2+1; end;
begin { Verify maximum (a,b,c)-value < 1.6 } Writeln(exp(2/exp(1)),' <',pow(1.6,1.6)); d := 1.6/veel/2; { half voxel size } min := 3; max := 0; { initialize } for i := 1 to veel-1 do begin for j := 1 to veel-1 do begin for k := 1 to veel-1 do begin ken := 0; { Binary number for collecting <> } { Each vertex of a 7-point star } vertex(-1,0,0); vertex(+1,0,0); vertex(0,-1,0); vertex(0,+1,0); vertex(0,0,-1); vertex(0,0,+1); if (ken = 0) or (ken = 63) then Continue; { Midpoint of star is near a^a*b^b*c^c = 1 } a := 2*i*d; b := 2*j*d ; c := 2*k*d; f := sqr(a)*b + sqr(b)*c + sqr(c)*a; if f < min then min := f; { Determine maximum of f(a,b,c) } if f > max then max := f; end; end; end; Writeln(min,' < f(a,b,c) <',max); end;
begin test(200); end.
And now we are curious, of course, what the maximum is (it's the last number in this output):

 2.08706522863453E+0000 < 2.12125057109759E+0000
 9.12537600000000E-0003 < f(a,b,c) < 3.00026265600000E+0000
Well, anyway better than the previous (Brute Force with Voxels) attempt. To be convincing, though, a decent error analysis is still needed :-(

Note. Explaining the estimate $\{a,b,c\} < \{1.6\}$ in the program: $$ f(x) = x^x = e^{x\ln(x)} \quad \Longrightarrow \quad f'(x) = [1+\ln(x)]e^{x\ln(x)} = 0 \quad \Longrightarrow \quad x=1/e \\ \Longrightarrow \quad f(1/e) = e^{-1/e} $$ This means that the maximum $x$ of each one of the coordinates in the product $a^ab^bc^c=1$ is: $$x^x = e^{2/e} < (1.6)^{1.6}$$ Therefore each of the coordinates $\;x < 1.6\,$ (or $\,1.58892154635044$ , to be double precise).

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  • $\begingroup$ A much simpler way to verify the inequality numerically (similar to what you have done for other inequality questions) is to check \eqref{*} instead: $$x^2y+y^2z+z^2x\le 3x^xy^yz^z$$ if $x+y+z=3$ and $x,y,z>0$. It is equivalent to the original inequality once they are homogenized. $\endgroup$ – Wiley Jul 18 '16 at 5:18
-1
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The condition can be presented in form $$a\ln a +b\ln b + c\ln c = 0.$$ Let $$a\leq b\leq c,$$ then $$\ln a \leq \ln b \leq \ln c,$$ and we can use Chebyshev sum inequality: $$3(a\ln a +b\ln b + c\ln c) \geq (a+b+c)\ln abc,$$ $$abc\leq 1.$$ Evidently, the expression $a^2b+b^2c+c^2a$ achieves maximum when $abc=1.$

The standard way of solving the problem on a conditional extremum is the method of Lagrange multipliers, which reduces it to a system of equations.

The maximum value of function $$f(a,b,c,λ)=a^2b+b^2c+c^2a+λ(abc-1)$$ for $a,b,c>0$ on the interval is reached or at its edges, or in one of the points with zero partial derivatives $$f'_a=0,\quad f'_b=0,\quad f'_c=0,\quad f'_λ=0,$$ or $$\begin{cases} 2ab+c^2+\lambda bc = 0\\ 2bc+a^2+\lambda ca = 0\\ 2ca+b^2+\lambda ab = 0\\ abc-1 = 0, \end{cases}$$ then $$\begin{cases} a(2ab+c^2) = b(2bc+a^2)=c(2ca+b^2),\\ abc=1, \end{cases}$$ $$\begin{cases} 2b^2c -c^2a = a^2b\\ b^2c -2c^2a = -a^2b\\ abc=1, \end{cases}$$ $$c^2a = b^2c = a^2b,\quad abc=1,$$ $$\dfrac cb =1, \dfrac ba =1, \quad abc=1,$$ $$a=b=c=1,$$ $$f(a,b,c,\lambda)=3.$$ Note that $$\lim_{a\to 0} a\ln a = 0,$$ so on the edge $a=0$ we have to minimize $f(b,c,\lambda) = b^2c + \lambda(bc-1),$ and then we get the system $$\begin{cases} 2bc+\lambda c = 0\\ b^2+\lambda b = 0\\ bc-1=0 \end{cases}$$ with solution $$b=2,\quad c=\dfrac12,\quad f = 2 < 3.$$

That means that $$\boxed{a^2b+b^2c+c^2a \leq 3}$$

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  • $\begingroup$ not sure why I still bother, but here it goes: you are repeating the exact same mistake in the most up-voted answer and it has been pointed out in the comment that $abc\le 1$ does NOT imply $a^2b+b^2c+c^2a\le 3$, take e.g., $(a,b,c)=(2,1/2,1)$ $\endgroup$ – Wiley Jul 26 '16 at 20:41

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