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

Solution 1:

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.

Solution 2:

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:

• $x+y^2+z^3=1$, prove $x^2y+y^2z+z^2x < \frac12$
• Olympiad Inequality $\sum_{cyc} \frac{x^4}{8x^3+5y^3} \geqslant \frac{x+y+z}{13}$

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 }

 1.00001285611974E+0000 < f < 1.68177794816992E+0000

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.