Kamran011 wrote: Let $x,y,z$ be positive real numbers such that $x^4+y^4+z^4=1$ . Determine with proof the minimum value of $\frac{x^3}{1-x^8}+\frac{y^3}{1-y^8}+\frac{z^3}{1-z^8}$ I'll write the solution later

@above, you need to check convexity before using Jensen

Kamran011 wrote: Let $x,y,z$ be positive real numbers such that $x^4+y^4+z^4=1$ . Determine with proof the minimum value of $\frac{x^3}{1-x^8}+\frac{y^3}{1-y^8}+\frac{z^3}{1-z^8}$ Azerbaijan 2019? Thanks.

sqing wrote: Kamran011 wrote: Let $x,y,z$ be positive real numbers such that $x^4+y^4+z^4=1$ . Determine with proof the minimum value of $\frac{x^3}{1-x^8}+\frac{y^3}{1-y^8}+\frac{z^3}{1-z^8}$ Azerbaijan 2019? Thanks. Yes, Azerbaijan Math Olympiad Training, I'll post other problems

Kamran011 wrote: Let $x,y,z$ be positive real numbers such that $x^4+y^4+z^4=1$ . Determine with proof the minimum value of $\frac{x^3}{1-x^8}+\frac{y^3}{1-y^8}+\frac{z^3}{1-z^8}$ Tangent line: $\frac{x^3}{1-x^8} \geq \frac{\sqrt[4]{3^9}}{8}x^4$

TL method works here Edit- sniped....

Math-wiz wrote: TL method works here Edit- sniped.... can you please send a link to this method?

Feridimo wrote: Math-wiz wrote: TL method works here Edit- sniped.... can you please send a link to this method? It is mentioned in Evan Chen's handout. Have a look. Its really helpful when Jensen dumps you!

Math-wiz wrote: Feridimo wrote: Math-wiz wrote: TL method works here Edit- sniped.... can you please send a link to this method? It is mentioned in Evan Chen's handout. Have a look. Its really helpful when Jensen dumps you! Agree. It was easy for Jensen

Math-wiz wrote: Feridimo wrote: Math-wiz wrote: TL method works here Edit- sniped.... can you please send a link to this method? It is mentioned in Evan Chen's handout. Have a look. Its really helpful when Jensen dumps you! How did you use Tangent Line method? Did you let $f(x) = \frac{\sqrt[4]{x^3}}{1-x^2}$ and then find the tangent to it at $x=\frac{1}{3}$? Is it true in general that if we find tangent for $g(x^r)$ as $h(x)$ then $h(\sqrt[r]{x})$ is tangent to $g(x)$? After substituting different power of the input variable ($x$) in a function what properties remain same and what changes?

Here's the TLT solution. It's cleaner than it looks, actually. The answer is $\boxed{\frac{27\sqrt[4]3}8}$. Let $a = x^4, b = y^4, c = z^4$. I claim that $$\sum \frac{a^{3/4}}{1-a^2} \geq \sum \frac {9 \sqrt[4] 3 a}8 = \frac{27\sqrt[4]3}8,$$with equality holding when $a=b=c=\frac 13$. Indeed, it suffices to verify the inequality $$\frac{x^3}{1-x^8} \geq \frac{9\sqrt[4]3}8 x^4 \iff x(1-x^8) \leq \frac 8{9\sqrt[4]3}.$$However, by AM-GM, we have $$\frac{8x^8+8(1-x^8)}9 \geq \sqrt[9]{(8x^8)(1-x^8)^8} \iff x^8(1-x^8)^8 \leq \frac{8^8}{9^9},$$which proves the result.

The expressions are one variable functions, we can use calculus to solve the problem, especially TLT. Firstyly, let $a=x^4,b=y^4,c=z^4$ in order to use TLT "better". Then Let $f(a)= \frac{a^{3/4}}{1-a^2}$. We wish to minimize $f(a)+f(b)+f(c)$ in subject to $a+b+c=1$. One may calculate $f'(a)=\frac{5a^2+3}{4a^{1/4}(1-a^2)^2}$. By TLT, there are some value $k$ such that $$ f(a) \ge f(k)+ f'(k)(a-k)$$One of the obvious values is $k=1/3$ which gives $$ f(a) \ge \frac{3^{5/4}}{8} + \frac{9(3^{1/4})}{8}(a-\frac{1}{3})$$ Then $$f(a)+f(b)+f(c) \ge \frac{27 \sqrt[4]{3}}{8}$$It is suffient to show that $\frac{x^3}{1-x^8} \ge \frac{9 \sqrt[4]{3}}{8}x^4 \Leftrightarrow 8 + x^9 3^{9/4} - 3^{9/4}x \ge 0$. Which is straight by calculus since derivative of LHS is $ 9\sqrt[4]{3}(9x^8-1)$ , which is $0$ when $x= \frac{1}{\sqrt[4]{3}}$. Hence we are done.

Wolfram Alpha here (see "factorization over splitting field") shows that if one looks for constants $A$ and $B$ such that $\frac{t^3}{1-t^8} \geq At^4 + B$ (with $t = 3^{-1/4}$ being a double root), one would get $B = 0$ and $A = \frac{9\sqrt[4]{3}}{8}$ and that the remaining multiplier after factoring has all its coefficients positive, hence it is positive itself. Thus in principle the linearization trick can be done by hand even without derivatives, but would require a bit of annoying pain.