When $a,b,c > 0$, prove $$\left(\frac{a^4}{a^3+b^3}\right)^{\frac34}+\left(\frac{b^4}{b^3+c^3}\right)^ {\frac34}+\left(\frac{c^4}{c^3+a^3}\right)^{\frac34} \geqslant \frac{a^{\frac34}+b^{\frac34}+c ^{\frac34}}{2^{\frac34}}$$
I tried the substitution $x=a^4,\ldots$ but I have no idea how to deal with the left- hand side. I tried some C-S but it goes nowhere. I think Bernoulli's inequality may be the only way to prove this inequality.
Let $x_1=a^3$, $x_2=b^3$, $x_3=c^3$, and $S=\sum_i x_i$. We wish to prove
$$ \frac{x_1}{(S-x_3)^{3/4}}+\frac{x_2}{{(S-x_1)}^{3/4}}+\frac{x_3}{{(S-x_2)}^{3/4}}-\frac{\sum_i x_i^{1/4}}{2^{3/4}}\ge 0.$$
We consider the problem to minimize $ \frac{x_1}{(S-x_3)^{3/4}}+\frac{x_2}{{(S-x_1)}^{3/4}}+\frac{x_3}{{(S-x_2)}^{3/4}}-\frac{\sum_i x_i^{1/4}}{2^{3/4}}$ over variables $x_1,x_2, x_3$, such that $x_i\ge 0$ and $\sum_i x_i = S$. We see that all $x_i$ equal satisfies KKT and is indeed the minimal choice (If Largrange multiplier for any of $x_i>0$ is non-zero, $x_i=0$ and the condition holds and thus we can assume that the Lagrange for each $x_i$ is zero and thus there is a single Lagrange multiplier for the sum and equal $x_i$ trivially satisfies the conditions.). Thus, we get the above expression to hold.
Define: $$ x = \frac{a^{3/4}}{2^{3/4}} = \left(\frac{a}{2}\right)^{3/4} \quad ; \quad y = \frac{b^{3/4}}{2^{3/4}} = \left(\frac{b}{2}\right)^{3/4} \quad ; \quad z = \frac{c^{3/4}}{2^{3/4}} = \left(\frac{c}{2}\right)^{3/4} $$ Then: $$ a = 2\,x^{4/3} \quad ; \quad b = 2\,y^{4/3} \quad ; \quad c = 2\,z^{4/3} $$ And: $$ \left(a^4\right)^{3/4} = a^3 = 2^3 x^4 \quad ; \quad \left(b^4\right)^{3/4} = b^3 = 2^3 y^4 \quad ; \quad \left(c^4\right)^{3/4} = c^3 = 2^3 z^4 $$ Finally, when $x,y,z > 0$, prove: $$ f(x,y,z) = \frac{2^3 x^4}{\left(2^3 x^4 + 2^3 y^4\right)^{3/4}} + \frac{2^3 y^4}{\left(2^3 y^4 + 2^3 z^4\right)^{3/4}} + \frac{2^3 z^4}{\left(2^3 z^4 + 2^3 x^4\right)^{3/4}} \ge 1 $$ Where it can be assumed without loss of generality that: $\,x+y+z = 1$ . The problem is thus reduced to a familiar one, quite similar to:
The minimum of our function inside the abovementioned triangle must shown to be greater or equal to one.
Another proof without words is attempted
by plotting a contour map of the function, as depicted. Levels (nivo) of these isolines are defined (in Delphi Pascal) as:
nivo := min + g/grens*(max-min); { grens = 20 ; g = 1..grens }
The whiteness of the isolines is proportional to the (positive) function values; they are almost black near the minimum and almost white near the maximum values.
Maximum and minimum values of the function are observed to be:
1.00001285611974E+0000 < f < 1.68177794816992E+0000The little $\color{blue}{\mbox{blue}}$ spot in the middle is where $\,\left| f(x,y,z) - 1 \right| < 0.001$ . Due to symmetry, an absolute minimum of the function is expected indeed at $(x,y,z) = (1/3,1/3,1/3)$.
Note. Conditions similar to $\;x+y+z=1\;$ often occur in these inequalities, whether that is explicitly or implicitly. An explicit example has been provided with another HN_NH question :
The current inequality is an example of implicit occurrence. Let the function $f(x,y,z)$ be defined as above, then we have the equivalent inequality $\;f(x,y,z) \geqslant x+y+z$ . It is clear that $f$ has the following property for all real $\lambda > 0$ : $\;f(\lambda x,\lambda y,\lambda z) = \lambda f(x,y,z) \geqslant \lambda (x+y+z)$ . Therefore $\lambda$ has no influence whatsoever on the inequality being true or false; we can always divide $x,y,z$ by a factor $\lambda$ such that $x+y+z=1$ . Thus enabling a triangle mapping method once again.
Version of 29.06.18
HINT
$$\mathbf{\color{brown}{Task\ transformations}}$$ Let $$ \begin{cases} &b^3+c^3= 2x^4\\ &c^3+a^3= 2y^4\\ &a^3+b^3= 2z^4 \end{cases}\Rightarrow \begin{cases} a^3=-x^4+y^4+z^4\\ b^3= x^4-y^4+z^4\\ c^3= x^4+y^4-z^4\tag1, \end{cases}$$ then the issue equation transforms to $$\rlap\bigcirc\!\sum\dfrac{-x^4+y^4+z^4}{z^3}\ge \rlap\bigcirc\!\sum\sqrt[4]{-x^4+y^4+z^4}.$$ $$\rlap\bigcirc\!\sum z\dfrac{-x^4+y^4+z^4}{z^4}\left(1-\dfrac1{\left(1+\dfrac{y^4-x^4}{z^4}\right)^{3/4}}\right)\ge 0.\tag2$$ Using inequality $$(1+t)^\alpha\le 1+\alpha t,\quad 1>\alpha>0,\quad t\ge-1,\tag3$$ for $\alpha=\dfrac34,$ can be obtained the stronger inequality than $(2):$ $$\rlap\bigcirc\!\sum z\left(1+\dfrac{y^4-x^4}{z^4}\right)\left(1-\dfrac1{1+\dfrac34\dfrac{y^4-x^4}{z^4}}\right)\ge 0,$$ $$\rlap\bigcirc\!\sum \dfrac{y^4-x^4}{z^3}\dfrac{z^4+y^4-x^4}{4z^4+3y^4-3x^4}\ge 0.\tag4$$
$$\mathbf{\color{brown}{Conditions}}$$
Taking in account $(1),$ the boundary conditions are $$0<x<\sqrt[4]{y^4+z^4},\quad 0<y<\sqrt[4]{z^4+x^4},\quad 0<z<\sqrt[4]{x^4+y^4}.\tag5$$
Taking in account $(5)$, term of $LSH(4)$ is negative iff $x > y.$ If $0<x\le y\le z,$ then the inequality $(4)$ is satisfied.
The task $(4)-(5)$ has rotational symmetry, so WLOG it is required to check only the case $$x\ge y\ge z>0.\tag6$$
The obtained task $(4)-(6)\ $ is correct and allows a simple proof.
It's not a proof, but I think it can help.
We can rewrite this inequality in the following form. $$\sum_{cyc}\frac{a^4}{\sqrt[4]{\left(\frac{a^4+b^4}{2}\right)^3}}\geq a+b+c,$$ where $a$, $b$ and $c$ are positives.
Now, since for $(a,b,c)=(1.98,0.89,1.38)$ $$\sum_{cyc}\frac{a^4}{\sqrt[4]{\left(\frac{a^4+b^4}{2}\right)^3}}-a-b-c=0.0080...,$$ we can use the Holder's inequality.
I checked that the following Holder does not help. $$\left(\sum_{cyc}\frac{a^4}{\sqrt[4]{(a^4+b^4)^3}}\right)^4\sum_{cyc}a^4(a^4+b^4)^3(a+mb+nc)^5\geq$$ $$\geq\left(\sum_{cyc}a^4(a+mb+nc)\right)^5$$ because the inequality $$8\left(\sum_{cyc}a^4(a+mb+nc)\right)^5\geq(a+b+c)^4\sum_{cyc}a^4(a^4+b^4)^3(a+mb+nc)^5$$ is wrong for all $m\geq0$ and $n\geq0$.
By the way, I think the following Holder can help.
$$\left(\sum_{cyc}\frac{a^4}{\sqrt[4]{(a^4+b^4)^3}}\right)^4\sum_{cyc}a^4(a^4+b^4)^3(a^2+kb^2+mc^2+lab+nac+pbc)^5\geq$$ $$\geq\left(\sum_{cyc}a^4(a^2+kb^2+mc^2+lab+nac+pbc)\right)^5,$$ where $a^2+kb^2+mc^2+lab+nac+pbc>0$ for all positives $a$, $b$ and $c$.
It's enough to prove that $$8\left(\sum_{cyc}a^4(a^2+kb^2+mc^2+lab+nac+pbc)\right)^5\geq$$ $$\geq(a+b+c)^4\sum_{cyc}a^4(a^4+b^4)^3(a^2+kb^2+mc^2+lab+nac+pbc)^5.$$ Now, we can substitute in the last inequality $(a,b,c)=(1.98,0.89,1.38)$
and we can choose parameteres $k,$ $l$, $m$, $n$ and $p$ such that the inequality is true.
I hope that it's possible! I think it's possible even for one or more of these parameters equal to zero. For this we need some software, which I have no.
After choosing of parameters we can try to prove the inequality by BW, for which we need software again.
About BW see here: https://artofproblemsolving.com/community/c6h522084
After proving of the starting inequality we can say that this inequality we can prove by hand, but after some days of idiotic computations.
Setting $a\rightarrow x^{1/3}$, $b\rightarrow y^{1/3}$, $c\rightarrow z^{1/3}$, the inequlity we want to show reduces to:
Show for any $x,y,z>0$ $$ \sum_{cyc}\frac{x}{(x+y)^{3/4}}\geq 2^{-3/4}\left(x^{1/4}+y^{1/4}+z^{1/4}\right).\tag 1 $$ We set $$ x=\frac{b_1+c_1-a_1}{2}\textrm{, }y=\frac{c_1+a_1-b_1}{2}\textrm{, }z=\frac{a_1+b_1-c_1}{2}, $$ where $a_1,b_1,c_1>0$ are the sides of a triangle and (1) becomes equivalent to $$ \sum_{cyc}\frac{\tau-a_1}{a_1^{3/4}}\geq 2^{-3/4}\left(\sum_{cyc}(\tau-a_1)^{1/4}\right),\tag 2 $$ where $\tau=\frac{a_1+b_1+c_1}{2}$. Hence equivalent we rewrite (2) in the form: $$ \left(\sum_{cyc}a_1\right)\left(\sum_{cyc}a_1^{-3/4}\right)-\left(\sum_{cyc}a_1^{1/4}\right)\geq \sum_{cyc}\left(a_1^{1/4}+(2\tau-2a_1)^{1/4}\right). $$ But from the "classical" inequality $$ x^{1/4}+y^{1/4}\leq 2^{3/4}\left(x+y\right)^{1/4}\textrm{, }\forall x,y>0, $$ we have (using that in every triangle we have $2\tau-2a_1$, $2\tau-2b_1$, $2\tau-2c_1>0$): $$ \sum_{cyc}a_1^{1/4}+\sum_{cyc}(2\tau-2a_1)^{1/4}\leq 2^{3/4} \sum_{cyc}(a_1+b_1)^{1/4}. $$ Hence if we manage to show that $$ 2^{3/4}\sum_{cyc}\left(a_1+b_1\right)^{1/4}+\sum_{cyc}a_1^{1/4}\leq\left(\sum_{cyc} a_1\right)\left(\sum_{cyc}a_1^{-3/4}\right) $$ we are done.
For this we use the Minkowski inequality:
If $n\in\{1,2,3,\ldots\}$ and $x_i,y_i\in\textbf{R}^{*}_{+}$, $k\in\textbf{R}^{*}_{+}$, then $$ \left[(x_1+y_1)^k+(x_2+y_2)^k+(x_3+y_3)^k\right]^{n-1}\leq $$ $$ \leq (x_1^k+x_2^k+x_3^k)^{n-1}+(y_1^k+y_2^k+y_3^k)^{n-1} $$ with equality when $\frac{x_1}{y_1}=\frac{x_2}{y_2}=\frac{x_3}{y_3}$.
We have (with $k=1/4$ and $n=5$): $$ 2^{3/4}\left((a_1+b_1)^{1/4}+(b_1+c_1)^{1/4}+(c_1+a_1)^{1/4}\right)\leq $$ $$ \leq 2^{3/4}\left[\left(a_1^{1/4}+b_1^{1/4}+c_1^{1/4}\right)^{n-1}+\left(b_1^{1/4}+c_1^{1/4}+a_1^{1/4}\right)^{n-1}\right]^{\frac{1}{n-1}}= $$ $$ =2\sum_{cyc}a_1^{1/4}. $$ From this it is clear that we only have to show: $$ 3\sum_{cyc}a_1^{1/4}\leq\left(\sum_{cyc}a_1\right)\left(\sum_{cyc}a_1^{-3/4}\right),\tag 3 $$ Which looks like Chebyshev inequality, but it is not.
A more general inequality of this kind is: $$ 3(x^{k-l}+y^{k-l}+z^{k-l})\leq (x^k+y^k+z^k)(x^{-l}+y^{-l}+z^{-l}),\tag 4 $$ where $x,y,z>0$ and $k>0$, $l>0$ and it is quite easy to prove. i.e.
$$ 3(x^{k-l}+y^{k-l}+z^{k-l})\leq (x^k+y^k+z^k)(x^{-l}+y^{-l}+z^{-l})\Leftrightarrow $$ $$ 2(x^{k-l}+y^{k-l}+z^{k-l})\leq (x^ky^{-l}+x^{-l}y^{k}+y^kz^{-l}+y^{-l}z^{k}+z^kx^{-l}+z^{-l}x^{k}). $$ But $$ x^{k-l}+y^{k-l}\leq x^{k}y^{-l}+y^{k}x^{-l}\Leftrightarrow $$ $$ x^{k}(x^{-l}-y^{-l})-y^{k}(-y^{-l}+x^{-l})\leq 0\Leftrightarrow $$ $$ (x^k-y^k)(x^{-l}-y^{-l})\leq0 $$ This last inequality is true for all $x,y>0$ and $k,l>0$. Gathering the other two inequalities involving $y,z$ and $z,x$, we get that (4) is true. Hence we get the validity of (3). QED
Partial answer :
Hint :
Using WRCF theorem see https://journalofinequalitiesandapplications.springeropen.com/articles/10.1186/1029-242X-2011-101
We have $c=\min{(a,b,c)}$ and let $f(x)$ such that :
$$f\left(x\right)=\left(\frac{1}{1+x^{3}}\right)^{\frac{3}{4}}$$
It seems we have $f''(x)>0$ for $x\geq 1$
For and $a,b,c\in[1,2]$ it seems we have :
$$g\left(x\right)=\frac{c^{\frac{3}{4}}}{a^{\frac{3}{4}}+b^{\frac{3}{4}}+c^{\frac{3}{4}}}f\left(x\right)+\left(1-\frac{c^{\frac{3}{4}}}{a^{\frac{3}{4}}+b^{\frac{3}{4}}+c^{\frac{3}{4}}}\right)f\left(\frac{\left(1-\frac{c^{\frac{3}{4}}}{a^{\frac{3}{4}}+b^{\frac{3}{4}}+c^{\frac{3}{4}}}x\right)}{1-\frac{c^{\frac{3}{4}}}{a^{\frac{3}{4}}+b^{\frac{3}{4}}+c^{\frac{3}{4}}}}\right)-f\left(1\right)\geq 0$$
Where $0.5\leq x\leq 1$ .
