$x,y,z >0$ and $x+y+z=3$, prove $$\tag{1}\frac{xy}{5y^3+4}+\frac{yz}{5z^3+4}+\frac{zx}{5x^3+4} \leqslant \frac13$$ My first attempt is to use Jensen’s inequality. Hence I consider the function $$f(x) =\frac{x}{5x^3+4}$$ Compute second derivative we have $$\tag{2}f”(x)=\frac{30x^2(5x^3-8)}{(5x^3+4)^3}$$ $(2)$ shows that the function is neither concave or convex. So I don’t think Jensen useful here. The author of $(1)$ is the same as of […]

Note: This question has a bounty that will expire in just a few days. Let $a,b,c$ and $d$ be the lengths of the sides of a quadrilateral. Show that $$ab^2(b-c)+bc^2(c-d)+cd^2(d-a)+da^2(a-b)\ge 0$$ Background: The well known 1983 IMO Problem 6 is the following: Let $ a$, $ b$ and $ c$ be the lengths of the […]

Let $a,b,$ and $c$ be the lengths of the sides of a triangle. Prove that $$\left | \dfrac{a-b}{a+b}+\dfrac{b-c}{b+c}+\dfrac{c-a}{c+a} \right | < \dfrac{1}{8}.$$ The best idea I had was to expand the fractions to get something nicer. So we get $\dfrac{a-b}{a+b}+\dfrac{b-c}{b+c}+\dfrac{c-a}{c+a} = \dfrac{a^2 b-a^2 c-a b^2+a c^2+b^2 c-b c^2}{(a+b) (a+c) (b+c)}.$ Then I would try to […]

$x,y,z >0$, prove $$\frac{x^4}{8x^3+5y^3}+\frac{y^4}{8y^3+5z^3}+\frac{z^4}{8z^3+5x^3} \geqslant \frac{x+y+z}{13}$$ Note: Often Stack Exchange asked to show some work before answering the question. This inequality was used as a proposal problem for National TST of an Asian country a few years back. However, upon receiving the official solution, the committee decided to drop this problem immediately. They don’t believe […]

For $x,y,z >0$, prove that $$\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y} \geqslant \sqrt{\frac94+\frac32 \cdot \frac{(y-z)^2}{xy+yz+zx}}$$ Observation: This inequality is stronger than the famous Nesbitt’s Inequality $$\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y} \geqslant \frac32 $$ for positive $x,y,z$ We have three variables but the symmetry holds only for two variables $y,z$, resulting in a very difficult inequality. Brute force and Largrange Multiplier are too complicated. The […]

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