USA IMO 2003

First Solution. (Based on work by Matthew Tang and Anders Kaseorg) By multiplying $a$, $b$, and $c$ by a suitable factor, we reduce the problem to the case when $a+b+c=3$. The desired inequality reads $$\frac{(a+3{)}^2}{2a^2+(3-a{)}^2}+\frac{(b+3{)}^2}{2b^2+(3-b{)}^2}+\frac{(c+3{)}^2}{2c^2+(3-c{)}^2}\le 8.$$ Set $$f(x)=\frac{(x+3{)}^2}{2x^2+(3-x{)}^2}$$ It suffices to prove that $f(a)+f(b)+f(c)\le 8$. Note that $$\begin{array}{rl}f(x)& =\frac{x^2+6x+9}{3(x^2-2x+3)}=\frac{1}{3}\cdot \frac{x^2+6x+9}{x^2-2x+3}\\ & =\frac{1}{3}\left(1+\frac{8x+6}{x^2-2x+3}\right)\\ & =\frac{1}{3}\left(1+\frac{8x+6}{(x-1{)}^2+2}\right)\le \frac{1}{3}\left(1+\frac{8x+6}{2}\right)\\ & =\frac{1}{3}(4x+4).\end{array}$$ Hence, $$f(a)+f(b)+f(c)\le \frac{1}{3}(4a+4+4b+4+4c+4)=8,$$ as desired, with equality if and only if $a=b=c$.

Second Solution. (By Liang Qin) Setting $x=a+b$, $y=b+c$, $z=c+a$ gives $2a+b+c=x+z$, hence $2a=x+z-y$ and their analogous forms. The desired inequality becomes $$\frac{2(x+z{)}^2}{(x+z-y{)}^2+2y^2}+\frac{2(z+y{)}^2}{(z+y-x{)}^2+2x^2}+\frac{2(y+x{)}^2}{(y+x-z{)}^2+2z^2}\le 8.$$ Because $2(s^2+t^2)\ge (s+t{)}^2$ for all real numbers $s$ and $t$, we have $2(x+z-y{)}^2+2y^2\ge (x+z-y+y{)}^2=(x+z{)}^2$. Hence $$\begin{array}{rl}\frac{2(x+z{)}^2}{(x+z-y{)}^2+2y^2}& =\frac{4(x+z{)}^2}{2(x+z-y{)}^2+4y^2}\le \frac{4(x+z{)}^2}{(x+z{)}^2+2y^2}\\ & =\frac{4}{1+2\cdot \frac{y^2}{(x+z{)}^2}}\le \frac{4}{1+2\cdot \frac{y^2}{2(x^2+z^2)}}\\ & =\frac{4(x^2+z^2)}{x^2+y^2+z^2}.\end{array}$$ It is not difficult to see that the desired result follows from summing up the above inequality and its analogous forms.

Third Solution. (By Richard Stong) Note that $$\begin{array}{rl}(2x+y{)}^2+2(x-y{)}^2& =4x^2+4xy+y^2+2x^2-4xy+2y^2\\ & =3(2x^2+y^2).\end{array}$$ Setting $x=a$ and $y=b+c$ yields $$(2a+b+c{)}^2+2(a-b-c{)}^2=3(2a^2+(b+c{)}^2).$$ Thus, we have $$\begin{array}{rl}\frac{(2a+b+c{)}^2}{2a^2+(b+c{)}^2}& =\frac{3(2a^2+(b+c{)}^2)-2(a-b-c{)}^2}{2a^2+(b+c{)}^2}\\ & =3-\frac{2(a-b-c{)}^2}{2a^2+(b+c{)}^2}.\end{array}$$ and its analogous forms. Thus, the desired inequality is equivalent to $$\frac{(a-b-c{)}^2}{2a^2+(b+c{)}^2}+\frac{(b-a-c{)}^2}{2b^2+(c+a{)}^2}+\frac{(c-a-b{)}^2}{2c^2+(a+b{)}^2}\ge \frac{1}{2}.$$ Because $(b+c{)}^2\le 2(b^2+c^2)$, we have $2a^2+(b+c{)}^2\le 2(a^2+b^2+c^2)$ and its analogous forms. It suffices to show that $$\frac{(a-b-c{)}^2}{2(a^2+b^2+c^2)}+\frac{(b-a-c{)}^2}{2(a^2+b^2+c^2)}+\frac{(c-a-b{)}^2}{2(a^2+b^2+c^2)}\ge \frac{1}{2},$$ or, $$(a-b-c{)}^2+(b-a-c{)}^2+(c-a-b{)}^2\ge a^2+b^2+c^2.$$ Multiplying this out, the left-hand side of the last inequality becomes $3(a^2+b^2+c^2)-2(ab+bc+ca)$. Therefore the last inequality is equivalent to $2[a^2+b^2+c^2-(ab+bc+ca)]\ge 0$, which is evident because $$2[a^2+b^2+c^2-(ab+bc+ca)]=(a-b{)}^2+(b-c{)}^2+(c-a{)}^2.$$ Equalities hold if and only if $(b+c{)}^2=2(b^2+c^2)$ and $(c+a{)}^2=2(c^2+a^2)$, that is, $a=b=c$.

Fourth Solution. We first convert the inequality into $$\frac{2a(a+2b+2c)}{2a^2+(b+c{)}^2}+\frac{2b(b+2c+2a)}{2b^2+(c+a{)}^2}+\frac{2c(c+2a+2b)}{2c^2+(a+b{)}^2}\le 5.$$ Splitting the 5 among the three terms yields the equivalent form $$\sum _{\text{cyc}}\frac{4a^2-12a(b+c)+5(b+c{)}^2}{3[2a^2+(b+c{)}^2]}\ge 0,(1)$$ where ${\sum }_{\text{cyc}}$ is the cyclic sum of variables $(a,b,c)$. The numerator of the term shown factors as $(2a-x)(2a-5x)$, where $x=b+c$. We will show $$\frac{(2a-x)(2a-5x)}{3(2a^2+x^2)}\ge -\frac{4(2a-x)}{3(a+x)}.(2)$$ Indeed, (2) is equivalent to $$(2a-x)[(2a-5x)(a+x)+4(2a^2+x^2)]\ge 0,$$ which reduces to $$(2a-x)(10a^2-3ax-x^2)=(2a-x{)}^2(5a+x)\ge 0,$$ which is evident. We proved that $$\frac{4a^2-12a(b+c)+5(b+c{)}^2}{3[2a^2+(b+c{)}^2]}\ge -\frac{4(2a-b-c)}{3(a+b+c)},$$ hence (1) follows. Equality holds if and only if $2a=b+c$, $2b=c+a$, $2c=a+b$, i.e., when $a=b=c$.

Fifth Solution. Given a function $f$ of $n$ variables, we define the symmetric sum $$\sum _{\text{sym}}f(x_1,\dots ,x_n)=\sum _{\sigma }f(x_{\sigma (1)},\dots ,x_{\sigma (n)})$$ where $\sigma$ runs over all permutations of $1,\dots ,n$ (for a total of $n!$ terms). For example, if $n=3$, and we write $x,y,z$ for $x_1,x_2,x_3$, $$\begin{array}{rl}\sum _{\text{sym}}{x}^3& =2x^3+2y^3+2z^3\\ \sum _{\text{sym}}{x}^{2}y& =x^{2}y+y^{2}z+z^{2}x+x^{2}z+y^{2}x+z^{2}y\\ \sum _{\text{sym}}xyz& =6xyz.\end{array}$$ We combine the terms in the desired inequality over a common denominator and use symmetric sum notation to simplify the algebra. The numerator of the difference between the two sides is $$2\sum _{\text{sym}}(4a^6+4a^{5}b+a^4{b}^2+5a^{4}bc+5a^3{b}^3-26a^3{b}^{2}c+7a^2{b}^2{c}^2),(3)$$ and it suffices to show the the expression in (3) is always greater or equal to 0. By the Weighted AM-GM Inequality, we have $4a^6+b^6+c^6\ge 6a^{4}bc$, $3a^{5}b+3a^{5}c+b^{5}a+c^{5}a\ge 8a^{4}bc$, and their analogous forms. Adding those inequalities yields $$\sum _{\text{sym}}6a^6\ge \sum _{\text{sym}}6a^{4}bc\text{and}\sum _{\text{sym}}8a^{5}b\ge \sum _{\text{sym}}8a^{4}bc.$$ Consequently, we obtain $$\sum _{\text{sym}}4a^6+4a^{5}b+5a^{4}bc\ge \sum _{\text{sym}}13a^{4}bc.(4)$$ Again by the AM-GM Inequality, we have $a^4{b}^2+b^4{c}^2+c^4{a}^2\ge 4a^2{b}^2{c}^2$, $a^3{b}^3+b^3{c}^3+c^3{a}^3\ge 3a^2{b}^2{c}^2$, and their analogous forms. Thus, $$\sum _{\text{sym}}{a}^4{b}^2+5a^3{b}^3\ge \sum _{\text{sym}}6a^2{b}^2{c}^2,$$ or $$\sum _{\text{sym}}{a}^4{b}^2+5a^3{b}^3+7a^2{b}^2{c}^2\ge \sum _{\text{sym}}13a^2{b}^2{c}^2.(5)$$ Recalling Schur's Inequality, we have $$\begin{array}{rl}& a^3+b^3+c^3+3abc-(a^{2}b+b^{2}c+c^{2}a+a{b}^2+b{c}^2+c{a}^2)\\ & =a(a-b)(a-c)+b(b-a)(b-c)+c(c-a)(c-b)\ge 0,\end{array}$$ or $$\sum _{\text{sym}}{a}^3-2a^{2}b+abc\ge 0.$$ Thus $$\sum _{\text{sym}}13a^{4}bc-26a^3{b}^{2}c+13a^2{b}^2{c}^2\ge 13abc\sum _{\text{sym}}{a}^3-2a^{2}b+abc\ge 0.(6)$$ Adding (4), (5), and (6) yields (3).