Vtyjkyuk

$x,y,z >0$, prove
$$fracx^xx+y+fracy^yy+z+fracz^zz+x geqslant frac32$$

I have a solution for this beautiful and elegant inequality. I am posting this inequality to share and see other solutions from everyone on MSE.

Since some of you asked for my solution, I used the following estimations
$$x^x geqslant frac12 left(x^2+1 right)$$

Proving rest of this inequality is very simple.

Method 1:
$$sum_cycfracx^xx+y geqslant frac12sum_cycfracx^2+1x+ygeqslant frac12 left( frac(x+y+z)^22(x+y+z)+sum_cycfrac1x+y right) geqslant frac12 sum_cycleft(fracx+y4+frac1x+y right) geqslant frac12 sum_cyc 2 cdotsqrtfracx+y4cdot frac1x+y geqslant frac32$$

Method 2:

$$sum_cyc fracx^xx+y geqslant frac12 sum_cycfracx^2+1x+y geqslant frac32 sqrt[3]frac(x^2+1)(y^2+1)(z^2+1)(x+y)(y+z)(z+x)geqslant frac32$$
It is because C-S yields $$(x^2+1)(y^2+1) geqslant (x+y)^2$$
so $$frac(x^2+1)(y^2+1)(z^2+1)(x+y)(y+z)(z+x) geqslant 1$$

Method 3:

Notice that
$$sum_cyc fracx^2+1x+y=sum_cyc fracy^2+1x+y$$
Hence
$$sum_cycfracx^xx+y geqslant frac12 sum_cycfracx^2+1x+y=frac14sum_cyc fracx^2+y^2+2x+y geqslantfrac14sum_cycfracfrac(x+y)^22+2x+y$$$$=frac14 sum_cyc left(fracx+y2+frac2x+y right) geqslant frac14 sum_cyc2 =frac32$$

Can you find a different solution that does not employ the estimation $x^x geqslant frac12(x^2+1)$ ?

Update 1

The solution of user260822 is not correct. This is not a Nesbitt's inequality.
We have
$$fracxx+y+fracyy+z+fraczz+x > 1 $$
(not $frac32$)

Update 2

Another similar question I posted a few weeks back

Unconventional Inequality $ fracx^xx-y+fracy^y+fracz^z > frac72$

Please help. Thank you

Since $e^xgeq 1+x$ for all $xin mathbbR$, by letting $x=-log u$ we have that for any $u >0$, $log ugeq 1-frac1u$, and so $ulog u>u-1$ for any $u>0$. Combining these two inequalities, it follows that $$sqrtx^x y^y= expleft(fracxlog x2+fracylog y2right)geq expleft(fracx+y2-1right)geq fracx+y2.$$ Hence $$fracx^xx+y+fracy^yy+z+fracz^zz+xgeq frac12 left(sqrtfracx^xy^y+sqrtfracy^yz^z+sqrtfracz^zx^xright),$$ and the result follows from the AM-GM inequality.

Remark: By combining $e^xgeq 1+x$ for $xinmathbbR$ and $ulog ugeq u-1$ for $u>0$ as above, we obtain the following "reverse exponential AM-GM inequality": $$sqrt[n]x_1^x_1x_2^x_2cdots x_n^x_ngeq fracx_1+x_2+cdots+x_nn.$$

Preliminary note: If $x=1+u$ and $y=1+v$ with some small $u,v$ then
$$
fracx^xx+y = frac(1+u)^1+u2+u+v approx frac12 frac1+u1+frac12u+frac12v approx frac12left(1+frac12u-frac12vright)
= frac12 + fracx-y4.
$$

We will prove that
$$
fracx^xx+y ge frac12 + fracx-y4. tag1
$$
Then the statement follows immediately.

(1) is equivalent with
$$
x^x ge fracx+y2 + fracx^2-y^24 = frac(x+1)^24 - frac(y-1)^24.
$$
The maximum of the RHS is attained at $y=1$, so it suffices to prove
$$
x^x ge frac(x+1)^24,
$$
or equivalently,
$$
f(x) = x ln x - ln4 -2ln(x+1)ge0.
$$

We have $f'(x) = 1+ln x - frac2x+1>0$ for $x>1$ and
$f'(x)<0$ for $0<x<1$.
Therefore, $min f = f(1)=0$ and $f(x)ge0$ for all $x>0$.
Equality occurs in (1) only for $x=y=1$.

An answer to your question ($x>0$): $enspacedisplaystyle x^x geqslant frac12(x^2+1)$ ?

Of course there are a lot of answers possible but here is an analytical one.

Claim: $enspacedisplaystyle sqrtx^sqrtx geqslant fracx+12$

We assume that we know that $,displaystyle x^x ,$
and therefore $,displaystyle sqrtx^sqrtx ,$ is convex.

Left an right side are equal for $x=1$ . So lets find the tangent $T(x)$ of $sqrtx^sqrtx$ in $(1,1)$ .

$displaystyle T(x)=(sqrtx^sqrtx)'|_x=1(x-1)+sqrtx^sqrtx|_x=1=(sqrtx^sqrtxfrac1+lnsqrtx2sqrtx)|_x=1(x-1)+1=fracx+12$

Because of the convexity of $sqrtx^sqrtx$ the claim follows.