1998 IMO Shortlist Problems/A3

Let $x,y,z$ be positive real numbers such that $xyz=1$ . Prove that $\frac{x^3}{(1+y)(1+z)}+\frac{y^3}{(1+x)(1+z)}+\frac{z^3}{(1+x)(1+y)}\geq\frac{3}{4}$

Solution1

Using Titu's Lemma, we rewrite the equation, getting that $\frac{(x^2+y^2+z^2)^2}{3xyz+2(xy+xz+yz)+x+y+z}\geq\frac{3}{4}$ . which means that $4(x^2+y^2+z^2)^2\geq 9+6(xy+yz+xz)+3(x+y+z)$ . Applying AM-GM inequality, we can get that $x^2+y^2+z^2\geq3(x^2y^2z^2)^\frac{1}{3}=3$ Which means the left side is greater than or equal to 36. Now let's observe the right side. Using C-S inequality we can easily get that $(x^2+y^2+z^2)^2\geq(xy+yz+xz)^2$ . So $2(x^2+y^2+z^2)^2\geq 6(xy+yz+xz)$ In the end, we can get that $(x^2+y^2+z^2)^2\geq(xy+yz+xz)^2=2xyz(x+y+z)+x^2y^2+x^2z^2+y^2z^2\geq 2xyz(x+y+z)+3(x^4y^4z^4)^\frac{1}{3}=2xyz+3$ Since $x^2+y^2+z^2\geq3$ , we can get that $x+y+z\leq3$ Now consider the right side $9+6(xy+yz+xz)+3(x+y+z)\leq9+6*3+3*3=36$ , the left side $4(x^2+y^2+z^2)^2\geq 4*9=36$ so left side is always larger or equal to right sides and we are done ~bluesoul

Solution2

WLOG, we assume that $x\leq y\leq z$ , so $\frac{1}{(1+y)(1+z)}\leq \frac{1}{(1+x)(1+z)}\leq \frac{1}{(1+x)(1+y)}$ . Now we can apply Chebyshev's inequality, getting that $\frac{x^3}{(1+y)(1+z)}+\frac{y^3}{(1+x)(1+z)}+\frac{z^3}{(1+x)(1+y)}\geq \frac{(x^3+y^3+z^3)(3+x+y+z)}{3(xy+xz+yz+x+y+z+1+xyz)}$ . Now we can observe that all three terms, $(x^3+y^3+z^3);(x+y+z);(xy+yz+xz)$ get their minimum value $3$ while $x=y=z$ . Which means the minimum value of the original expression holds while $x=y=z=1$ getting $\frac{3}{4}$ and we are done ~bluesoul

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1998 IMO Shortlist Problems/A3

Solution1

Solution2