# Prove $\sum\limits_{cyc}\left(\frac{a^4}{a^3+b^3}\right)^{\frac34} \geqslant \frac{a^{\frac34}+b^{\frac34}+c^{\frac34}}{2^{\frac34}}$

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.

• Would you mind telling me where you get your inequalities from? May 21, 2016 at 2:01
• can you post us this material? May 21, 2016 at 17:53
• Did you try holder's inequality? May 28, 2016 at 17:08
• After rewriting of the inequality in the form $\sum\limits_{cyc}\frac{a^4}{\left(\frac{a^4+b^4}{2}\right)^{\frac{3}{4}}}\geq a+b+c$ we get a big problem around (0.8372,1.2954,1.8596). I think it's impossible to prove this inequality without computer. Sep 24, 2016 at 7:16
• @HN_NH I proved tens of thousands of inequalities. How many inequalities you proved? I did not see even one proof from you. Sep 29, 2016 at 3:55

## 6 Answers

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.

• It is fairly obvious that $x_1=x_2=x_3$ is an extremal point, but how do you know that this is a global minimum? This is the real problem when applying this method and requires you to solve the equations defining the extremal points (or proving that they do not have any other solutions). Jun 1, 2016 at 5:38
• Convexity of the expression. Jun 1, 2016 at 5:41
• Yes that should be enough. One still needs to show or argue why the expression is convex though (it might be obvious, but I don't see a simple way to see this). Jun 1, 2016 at 5:52
• Product of convex, non-negative, non-decreasing functions is convex (extendible for multi-variate case - math.stackexchange.com/questions/27571/… ). $x_1$ and $1/(S-x_3)^{3/4}$ satisfy - product is convex - sum is convex, $-x_i^{1/4}$ is convex, so overall expression is easily shown to be multi-variate convex. Jun 1, 2016 at 5:58
• How do you define monotonicity for a function with more than one variable? The function is monotone if $S$ is fixed, which is not the case. The convexity should be then the whole convexity. Jun 1, 2016 at 13:44

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:

And can be treated accordingly:

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+0000

The 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 :

• Prove $\frac{xy}{5y^3+4}+\frac{yz}{5z^3+4}+\frac{zx}{5x^3+4} \leqslant \frac13$
• 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.

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.

• (+1) Your discussion is helpful. Jun 3, 2020 at 14:05
• +1, But how do you know $(a,b,c)=(1.98,0.89,1.38)$? Sep 15, 2020 at 13:16
• @tthnew I just tried to prove it and found that there is this strange point. Sep 15, 2020 at 13:20

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.

• You said stronger inequality than (2). I think that it it not true. (4) is weaker than (2). Since $(1 + \frac{y^4-x^4}{z^4})^{3/4} \le 1 + \frac{3}{4}\frac{y^4-x^4}{z^4}$ (according to (3)), actually you increase each term in (2). Please check it. Jun 5, 2020 at 13:54

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

• If you need anything to ask about my proof please be free to. Jun 30, 2018 at 13:58
• I think that (1) is not equivalent to (2). We have $c_1 = x + y, b_1 = z+x, a_1 = y+z$ and $\tau - a_1 = x$, $\tau - b_1 = y, \tau - c_1 = z$. So, $\frac{\tau - a_1}{a_1^{3/4}} = \frac{x}{(y+z)^{3/4}}$ which is different from $\frac{x}{(x+y)^{3/4}}$. In other words, we need to prove $\sum_{\mathrm{cyc}} \frac{x}{(x+y)^{3/4}} \ge 2^{-3/4}(x^{1/4} + y^{1/4} + z^{1/4})$, but you go to prove $\sum_{\mathrm{cyc}} \frac{x}{(y+z)^{3/4}} \ge 2^{-3/4}(x^{1/4} + y^{1/4} + z^{1/4})$. By the way, actually, you prove a weaker version of the original inequality. Jun 5, 2020 at 16:06

Partial answer :

Hint :

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 $$0.5\leq x\leq 1$$ .

• @RiverLi We are near from the hard case no ? Apr 5, 2022 at 8:37