For any real numbers $x , y ,z$ prove that : $$(x+y+z)^2 + \sum_{cyc}{\frac{(x+y)(y+z)}{1+|x-z|}} \ge xy+yz+zx$$ Proposed by Navid Safaei
Problem
Source: Iran Team selection test 2024 - P3
Tags: algebra, inequalities
19.05.2024 19:01
Let $2a=y+z, 2b=z+x, 2c=x+y$. Then we can rewrite our inequality how $$a^2+b^2+c^2+\sum \frac{2ab}{1+2|a-b|} \geqslant 0. (*)$$WLOG $ab \geqslant 0$ and $a \geqslant b \geqslant c$ ("function" of $a,b,c$ (in LHS) is even, so we can assume this). Also let $2(a-b)=k, 2(b-c)=n \Rightarrow 2(a-c)=k+n$. Then we need to prove that $c^2+2c\left (\frac{b}{1+n}+\frac{a}{1+k+n} \right ) + \left ( a^2+b^2+\frac{2ab}{1+k} \right ) \geqslant 0$. But $\Delta_c/4=(b/(1+n)+a/(1+k+n))^2-(a^2+b^2+2ab/(1+k)) \leqslant 0$, because $a, b, n, k \geqslant 0$ (after this it is obviously). So, it is end of proof. P.S. for making below proof right note that we should only replace $a \to a/2, b \to b/2, c \to c/2$ in inequality (*), as was in first version of my solution.
19.05.2024 19:06
Like above, it's enough to prove that $$\sum a^2 + 2\sum\frac{ab}{1+| a-b|}\geq 0,$$for all real numbers $a,b,c$. Take $a,b$ two real numbers. If $a,b$ have opposite signs, $|a-b|=|a|+|b|$. If they have the same sign, quite clearly (by the triangle inequality) $|a-b|\leq |a|+|b|$. So anyways, $$\frac{ab}{1+|a-b|}=ab\int_0^1x^{|a-b|}dx\geq ab\int_0^1 x^{|a|+|b|}dx=\int_0^1 ax^{|a|}\cdot bx^{|b|}dx,$$for any $a,b\in\mathbb{R}$. Thus, $$\sum a^2 + 2\sum \frac{ab}{1+|a-b|}\geq \int_0^1 \left(\sum ax^{|a|}\right)^2 dx\geq 0.$$Of course, equality if and only if $a=b=c=0$. @below, if $ab\leq 0$, the inequality is actually an equality. If $ab$ is positive, since $x\in (0,1)$, we have $x^{|a-b|}\geq x^{|a|+|b|}$.
19.05.2024 23:26
I understand now, thanks.
16.08.2024 16:50
Do the same substitution and assume $a\ge b$. Clearly, we only need to check what happens when one of $a,b,c$ say $c$ is negative. Replace $c$ by $-c$ and expand: $8a^4b+8a^4c+4a^4+4a^3b+8a^3c^2+12a^3c+4a^3+8a^2b^3+8a^2b^2+12a^2bc+4a^2b+8a^2c^3+a^2+4ab^3+36ab^2c+8ab^2+12abc^2+2ab+8ac^4+12ac^3+8b^2c^+8b^2c+b^2+4c^4+4c^3+c^2\ge 12a^2bc+8ab^4+2ac+8b^3c+4b^4+8b^3c^2+4b^3c+8b^2c^3+8bc^4+4bc^3+4bc^2+2bc$ $8a^4b\ge 8a^3b^2$, $12a^3c\ge12a^2bc$, $8a^2b^3\ge 8ab^4$, $8a^4c\ge 8b^4c$, $4a^4\ge 4b^4$, $8a^3c^2\ge 8b^3c^2$, $4ab^3c\ge 4b^3c$, $8a^2c^3\ge 8b^2c^3$, $8ac^4\ge 8bc^4$, $4ac^3\ge 4bc^3$, $2a^34c^3\ge 4 bc^2$ and finally $c^2+(b+a)^2\ge 2ac+2cb$