USA IMO

First Solution. (By Richard Stong) From the condition, at least one of $a,b$, and $c$ does not exceed $1$, say $a\le 1$. Then $$ab+bc+ca-abc=a(b+c)+bc(1-a)\ge 0.$$ To obtain equality, we have $a(b+c)=bc(1-a)=0$. If $a=1$, then $b+c=0$ or $b=c=0$, which contradicts the given condition $a^2+b^2+c^2+abc=4$. Hence $1-a\ne 0$ and only one of $b$ and $c$ is $0$. Without loss of generality, say $b=0$. Therefore $b+c>0$ and $a=0$. Plugging $a=b=0$ back into the given condition gives $c=2$. By permutation, the lower bound holds if and only if $(a,b,c)$ is one of the triples $(2,0,0)$, $(0,2,0)$, and $(0,0,2)$.

Now we prove the upper bound. Let us note that some two of the three numbers $a,b$, and $c$ are both greater than or equal to $1$ or less than or equal to $1$. Without loss of generality, we assume that the numbers with this property are $b$ and $c$. Then we have $$(1-b)(1-c)\ge 0.(1)$$ The given equality $a^2+b^2+c^2+abc=4$ and the inequality $b^2+c^2\ge 2bc$ imply $$a^2+2bc+abc\le 4,\text{or}bc(2+a)\le 4-a^2.$$ Dividing both sides of the last inequality by $2+a$ yields $$bc\le 2-a.(2)$$ Combining (1) and (2) gives $$\begin{array}{rl}ab+bc+ac-abc& \le ab+2-a+ac(1-b)\\ & =2-a(1+bc-b-c)\\ & =2-a(1-b)(1-c)\le 2,\end{array}$$ as desired.

The last equality holds if and only if $b=c$ and $a(1-b)(1-c)=0$. Hence, equality for the upper bound holds if and only if $(a,b,c)$ is one of the triples $(1,1,1)$, $(0,\sqrt{2},\sqrt{2})$, $(\sqrt{2},0,\sqrt{2})$, and $(\sqrt{2},\sqrt{2},0)$.

Second Solution. (by Oaz Nir) We prove only the upper bound here. Either two of $a,b,c$ are less than or equal to $1$, or two are greater than or equal to $1$. Assume $b$ and $c$ have this property. Then $$b+c-bc=1-(1-b)(1-c)\le 1.(3)$$ Viewing the given equality as a quadratic equation in $a$ and solving for $a$ yields $$a=\frac{-bc\pm \sqrt{(b^2-4)(c^2-4)}}{2}.$$ Note that $$(b^2-4)(c^2-4)=b^2{c}^2-4(b^2+c^2)+16$$ $$\le b^2{c}^2-8bc+16=(4-bc{)}^2.$$ For the given equality to hold, we must have $b,c\le 2$ so that $4-bc\ge 0$. Hence, $$a\le \frac{-bc+|4-bc|}{2}=\frac{-bc+4-bc}{2}=2-bc,$$ or $$2-bc\ge a.(4)$$ Combining (3) and (4) gives $$2-bc\ge a(b+c-bc)=ab+ac-abc,$$ or $$ab+ac+bc-abc\le 2,$$ as desired.

Third Solution. We prove only the upper bound here. Define functions $f,g$ as $$f(x,y,z)=x^2+y^2+z^2+xyz=(x+y{)}^2+z^2-(2-z)xy,$$ $$g(x,y,z)=xy+yz+zx-xyz=z(x+y)+(1-z)xy$$ for all nonnegative numbers $x,y,z$. Observe that if $z\le 1$, then both $f$ and $g$ are unbounded, increasing functions of $x$ and $y$.

Assume that $f(a,b,c)=4$ and, without loss of generality, that $a\ge b\ge c\ge 0$. Then $c\le 1$. Let $a^{\prime }=(a+b)/2$. Because $a+b=a^{\prime }+a^{\prime }$ and $ab\le (\frac{a-b}{2}{)}^2+ab=a^{\prime 2}$, we have $$f(a^{\prime },a^{\prime },c)\le f(a,b,c)=4\text{and}g(a^{\prime },a^{\prime },c)\ge g(a,b,c).$$ Now increase $a^{\prime }$ to $e\ge 0$ such that $f(e,e,c)=4$. Note that $g(e,e,c)\ge g(a^{\prime },a^{\prime },c)$. It suffices to prove that $g(e,e,c)\le 2$.

Since $f(e,e,c)=2e^2+c^2+e^{2}c=4$, $e^2=(4-c^2)/(2+c)=2-c$. We obtain that $$\begin{array}{rl}g(e,e,c)-2& =2ec+(1-c)e^2-2\\ & \le 2e(2-e^2)+(e^2-1)e^2-2\\ & =e^4-2e^3-e^2+4e-2\\ & =(e-1{)}^2(e^2-1)\le 0,\end{array}$$ since $1\le e\le \sqrt{2}$.

Fourth Solution. We prove only the upper bound here. It is clear that $a,b,c\le 2$. If $a=0$, then the given equality reduces to $b^2+c^2=4$. Then $ab+bc+ca-abc=bc\le \frac{(b-c{)}^2}{2}+bc=\frac{b^2+c^2}{2}=2$. Similarly, the inequality is true if $b$ or $c$ equals $0$.

Suppose that $a,b,c>0$. Solving for $a$ yields $$\begin{array}{rl}a& =\frac{-bc+\sqrt{b^2{c}^2-4(b^2+c^2-4)}}{2}\\ & =\frac{-bc+\sqrt{(4-b^2)(4-c^2)}}{2}.\end{array}$$ We make the trigonometric substitution $b=2\sin u$ and $c=2\sin v$, where $0^{\circ }<u,v<90^{\circ }$. Then $$a=2(-\sin u\sin v+\cos u\cos v)=2\cos (u+v).$$ Set $u=B/2$, $v=C/2$, and $A=180^{\circ }-B-C$. Then $a=2\cos (u+v)=2\sin (A/2)$, $b=2\sin (B/2)$, and $c=2\sin (C/2)$, where $A,B,C$ are the angles of a triangle. We have $$\begin{array}{rl}ab& =4\sin \frac{A}{2}\sin \frac{B}{2}=2\sqrt{\sin A\tan \frac{A}{2}\sin B\tan \frac{B}{2}}\\ & =2\sqrt{\sin A\tan \frac{B}{2}\sin B\tan \frac{A}{2}}\end{array}$$ By the AM-GM Inequality, this is at most $$\begin{array}{rl}\sin A\tan \frac{B}{2}+\sin B\tan \frac{A}{2}& \\ & =\sin A\cot \frac{A+C}{2}+\sin B\cot \frac{B+C}{2}.\end{array}$$ Likewise, $$bc\le \sin B\cot \frac{B+A}{2}+\sin C\cot \frac{C+A}{2},$$ $$ca\le \sin C\cot \frac{C+B}{2}+\sin A\cot \frac{A+B}{2}.$$ Therefore, applying the Sum-to-product, Product-to-sum, and Double-angle formulas, we have $$\begin{array}{rl}ab+bc+ca& \le (\sin A+\sin B)\cot \frac{A+B}{2}+(\sin B+\sin C)\cot \frac{B+C}{2}\\ & +(\sin C+\sin A)\cot \frac{C+A}{2}\\ & =2\cos \frac{A-B}{2}\cos \frac{A+B}{2}+2\cos \frac{B-C}{2}\cos \frac{B+C}{2}\\ & +2\cos \frac{C-A}{2}\cos \frac{C+A}{2}\\ & =2(\cos A+\cos B+\cos C)\\ & =6-4\left({\sin }^2\frac{A}{2}+{\sin }^2\frac{B}{2}+{\sin }^2\frac{C}{2}\right)\\ & =6-(a^2+b^2+c^2).\end{array}$$ Using the given equality, this last quantity equals $2+abc$. It follows that $$ab+bc+ca\le 2+abc,$$ as desired.