I think there is no maxima $a\to\infty, b\to 1, c=5$ gives contradiction Edit- the problem was corrected later

Math-wiz wrote: I think there is no maxima $a\to\infty, b\to 1, c=5$ gives contradiction Yes! Thanks! I just copied and pasted how it was sent to me. It looks that there is a mistake in translation from Russian into English

“bigant146” did you attend this exam?

On a second note, I think this problem must be having a better solution. Can someone send the official solution? @4below wow nice solution

Let real $a$, $b$, and $c$ satisfy $abc+a+b+c=ab+bc+ca+5.$ Then $$a^2+b^2+c^2\geq 6.$$Equality holds when$\{a,b,c\}=\{2,-1,-1\}.$

sqing wrote:

Let real $a$, $b$, and $c$ satisfy $abc+a+b+c=ab+bc+ca+5.$ Then $$a^2+b^2+c^2\geq 6.$$Equality holds when$\{a,b,c\}=\{1,-2,-2\}.$ I made a mistake in typing, Sir. Equality will hold when $(x,y,z)=(-2,-2,1)$ so $(a,b,c)=(-1,-1,2)$

$$\iff$$Let $a$, $b$, and $c$ be reals such that $abc=4.$ Prove that$$(a+1)^2+(b+1)^2+(c+1)^2\geq 6.$$Equality holds when$\{a,b,c\}=\{1,-2,-2\}.$

This is the solution I've made in the exam.

abc+a+b+c=ab+bc+ca+5,equivalent to:(1-a)(1-b)(1-c)=-4,let:1-a=x,1-b=y,1-c=z,rezult:xyz=-4 and a^2+b^2+c^2=(1-x)^2+(1-y)^2+(1-z)^2= =2-2(x+y)+x^2+y^2+(1+4/xy)^2=2-2s+s^2-2p+(1+4/p)^2,where s=x+y,p=xy,let fp)=2-2s+s^2-2p+(1+4/p)^2.f'(p)<0,f=decreasing,rezult f(p)>=f(s^2/4)= =2-2s+s^2-s^2/2+(1+16/s^2)^2=g(s) and g'(4)=0,g"(s)>0,rezult:g(s)>=g(4)=2-8+8+4=6,the minimum is 6

Let $a\ge b\ge c$, then from $(a-1)(b-1)(c-1)=4\Longrightarrow a-1>0$ so $a> 1$. \begin{align*} a^2+b^2+c^2=&a^2-2+(b^2+c^2+1+1)\ge a^2-2+\frac{(b+c-1-1)^2}{4}\ge a^2-2+(b-1)(c-1)=a^2-2+\frac{4}{a-1}\\ =&(a-1)^2+2(a-1)+\frac{4}{a-1}-1\ge7\sqrt[7]{(a-1)^2(a-1)(a-1)\frac{1}{a-1}\frac{1}{a-1}\frac{1}{a-1}\frac{1}{a-1}}-1=6 \end{align*}

sqing wrote: Let $a$, $b$, and $c$ be reals such that $abc=4.$ Prove that$$(a+1)^2+(b+1)^2+(c+1)^2\geq 6.$$Equality holds when$\{a,b,c\}=\{1,-2,-2\}.$

sqing wrote: Let real $a$, $b$, and $c$ satisfy $abc+a+b+c=ab+bc+ca+5.$ Then $$a^2+b^2+c^2\geq 6.$$Equality holds when$\{a,b,c\}=\{2,-1,-1\}.$