Trivial by $s = a + b$ and $p = ab$. We have $s^3 - 3sp = p + 1$, so $p = \frac{s^3-1}{3s + 1}$ (we cannot have $s = -\frac{1}{3}$, as then $s^3 - 1 \neq 0$). From $p \leq \frac{s^2}{4}$ (equivalent to $(a-b)^2 \geq 0$), we get $s^3 - s^2 - 4 \leq 0$, i.e. $(s-2)(s^2 + s + 2) \leq 0$, so $s \leq 2$. We wish to prove $s^2 - 4p + s \geq 2$, i.e. $p \leq \frac{s^2 + s - 2}{4}$. Equivalently, $\frac{s^3-1}{3s+1} \leq \frac{s^2+s-2}{4}$, which rearranged to $s^3 - 4s^2 + 5s - 2 \leq 0$, factoring as $(s-2)(s+1)^2 \leq 0$, true by the abovementioned $s\leq 2$.

Iveela wrote: Let $a, b$ be positive reals such that $a^3 + b^3 = ab + 1$. Prove that \[(a-b)^2 + a + b \geq 2\] See here

arqady wrote: Iveela wrote: Let $a, b$ be positive reals such that $a^3 + b^3 = ab + 1$. Prove that \[(a-b)^2 + a + b \geq 2\] See here Yeah, I saw that the problem was somehow leaked during the competition, better to ignore/delete that post.

"I’m not leaked. I just saw Izho’s problems (my friend sent them to me), and I thought the time was over, but a guy told me it wasn’t. then I tried to delete my two posts." ( my fault)

Titibuuu wrote: "I’m not leaked. I just saw Izho’s problems (my friend sent them to me), and I thought the time was over, but a guy told me it wasn’t. then I tried to delete my two posts." ( my fault) Ok, no worries, accidents happen. Can you please post Problem 2 again? (it is the only one from today currently not posted)

t Assassino9931 wrote: Titibuuu wrote: "I’m not leaked. I just saw Izho’s problems (my friend sent them to me), and I thought the time was over, but a guy told me it wasn’t. then I tried to delete my two posts." ( my fault) Ok, no worries, accidents happen. Can you please post Problem 2 again? (it is the only one from today currently not posted) Thank you for understanding me, and I'm sorry again. I will post now

I apologize if the Olympiad in which Izho participated gave the impression that it was unfair, and I hope that nothing like this has happened

It seems there are $2$ posts with this problem. Here is the solution I've posted on the other one as well: By AM-GM we have $ab+2=a^3+b^3+1\geq 3ab\implies ab\leq 1$ $a^3+b^3=ab+1\implies (a+b)(a^2-ab+b^2)=ab+1\implies a^2-ab+b^2=\dfrac{ab+1}{a+b}$ $(a-b)^2+a+b\geq 2\iff a^2+b^2-ab+a+b-ab\geq 2\iff \dfrac{ab+1}{a+b}+a+b-ab\geq 2\iff$ $\iff (a+b)^2-ab(a+b)+ab+1\geq 2(a+b)\iff (a+b-1)^2+ab(1-a-b)\geq 0\iff$ $\iff -(a-1)(b-1)(a+b-1)\geq 0\iff (a-1)(b-1)(a+b-1)\leq 0$ If $a\geq 1\implies b\leq 1$ (since $ab\leq 1$) and $a+b>1$, so the inequality holds. Similarly, if $b\geq 1$ the inequality holds. Claim: The equation $a^3+b^3=ab+1$ doesn't have any solutions in $(0,1)^2$. Proof: Suppose there exists $(a,b)\in(0,1)^2$. $a, b<1\implies a^3<a$ and $b^3<b$, so $1=a^3+b^3-ab<a+b-ab\implies ab-a-b+1<0\implies (a-1)(b-1)<0$, which is clearly false and we're done.

Here is my solution from a thread earlier. Apparently #2 is literally same as this. Let $a+b=u$ and $ab=v$ ($u,v \in \mathbb{R^{+}}$ since $a,b \in \mathbb{R^{+}}$) By AM-GM Inequality, $$a^3+b^3+1 \geq 3ab$$$$ \implies ab+2 \geq 3ab \implies \boxed{v \leq 1} $$By Power-Mean Inequality, $${\left(\frac{a^3+b^3+1}{3}\right)}^{1/3} \geq \left(\frac{a+b+1}{3}\right)$$$$\implies {\left(\frac{ab+2}{3}\right)}^{1/3} \geq \left(\frac{a+b+1}{3}\right) \implies \frac{u+1}{3} \leq {\left(\frac{v+2}{3}\right)}^{1/3} \leq 1(\because v \leq 1)$$$$ \implies \boxed{u \leq 2}$$Now, $a^3+b^3=ab+1 \implies (a+b)^3-3ab(a+b)=(ab+1) \implies u^3-3uv=v+1 \implies \boxed{v=\frac{u^3-1}{3u+1}}$. We wish to prove that $$ (a-b)^2+a+b \geq 2 $$$$ \iff (a+b)^2+(a+b)-4ab \geq 2 \iff u^2+u-4v \geq 2$$$$ \iff (u^2+u)-4\left(\frac{u^3-1}{3u+1}\right) \geq 2$$$$ \iff (u^2+u)(3u+1)-4(u^3-1) \geq 2(3u+1)$$$$ \iff -u^3+4u^2-5u+2 \geq 0 \iff \boxed{(u-1)^{2}(2-u) \geq 0}$$which is indeed true. $\blacksquare$ ($\mathcal{QED}$)

Here is the shortest proof: $$a^2+b+b^2+a\ge 2\sqrt{(a^2+b)(b^2+a)}=2(ab+1)$$

Let $ a,b\geq 0 $ and $ a^3 + b^3 - ab\geq 1.$ Prove that $$(a - b)^2 + a + b \geq 2$$

rightways wrote: Here is the shortest proof: $$a^2+b+b^2+a\ge 2\sqrt{(a^2+b)(b^2+a)}=2(ab+1)$$ Very nice.

Can we use vieta formulas in this problem ? I try to find alternative solution

let $f(a,b)=a^2-2ab+a+b^2+b-2$ and $g(a,b)=a^3-ab+b^3-1$. now we can use the method of Lagrange multipliers; in the minimum point of $f$ we should have: $ \nabla f = \lambda \nabla g \Longleftrightarrow 2a-2b+1 = \lambda (3a^2 - b) \quad \& \quad 2b-2a+1 = \lambda (3b^2-a)$. if $3b^2 = a$ then $a = b + \frac{1}{2}$ so $3b^2 - b - \frac{1}{2} = 0$ but this equation doesn't have any real root ,contradiction!. so we can write equations in this form: $\frac{2a-2b+1}{3a^2-b}=\frac{2b-2a+1}{3b^2-a} \Longleftrightarrow (a-b)(6a^2+6b^2-5a-5b-1)=0$ in the case $a=b=x$ we have $2x^3-x^2-1=(x-1)(2x^2 + x + 1)=0$ the second term doesn't have positive root so we get $a=b=1$ and it's easy to see they work! so we can assume that $h(a,b) = 6a^2+6b^2-5a-5b-1 = 0$. now we found out in minimum point of $f$ with knowing $g=0$ then $h=0$. we can prove minimum of $f$ on $h$ is bigger than 0 and it proves the problem!!. for this we use the method of Lagrange multipliers again!!. $ \nabla f = \lambda \nabla h \Longleftrightarrow 2a-2b+1 = \lambda (12a-5) \quad \& \quad 2b-2a+1 = \lambda (12b-5)$ so $4(a-b) = 12\lambda (a-b)$ then with knowing $a \neq b$ we have $\lambda = \frac{1}{3}$ and we get $2b-2a+1 = 4b - \frac{5}{3} \Longleftrightarrow a+b = \frac{4}{3}$ applying on $h=0$ we get: $h(a,b,c)= 6(a+b)^2 -12ab - 5(a+b) -1 = 3 - 12ab = 0 \Longleftrightarrow ab = \frac{1}{4}$. so we have to prove $f$ is positive on the point $(a,b)$ and it's: $f(a,b) = (a+b)^2 - 4ab +(a+b) - 2 = \frac{1}{9} > 0$. and the problem is done!

Really easy

Let $ a, b $ be reals such that $ a^2+b^2+ a^5+ b^5 = ab + 3 $. Prove that\[(a-b)^2 + a + b \geq 2\]Let $ a, b $ be reals such that $ a^3 + b^3+ a^5+ b^5 = ab +3 $. Prove that\[(a-b)^2 + a + b \geq 2\]Let $ a, b $ be reals such that $ a^2+b^2+a^3 + b^3+ a^5+ b^5 = ab + 5 $. Prove that\[(a-b)^2 + a + b \geq 2\]h