[nooonuii]

คิดข้อ 7 ผิดครับ งั้นขอเฉลยสามข้อนี้ก่อนละกัน

31,87,92 สามข้อนี้สามารถพิสูจน์ได้ด้วยอสมการข้อ 87 ครับ

87. USA 1997 $(a,b,c>0)$
$$\frac{1}{a^3+b^3+abc}+\frac{1}{a^3+b^3+abc}+\frac{1}{a^3+b^3+abc}\leq \frac{1}{abc}$$

Solution :

We can easily show that
$$a^3+b^3\geq ab(a+b).$$
Thus $$\frac{1}{a^3+b^3+abc}\leq \frac{1}{ab(a+b+c)}$$
Hence
$$LHS \leq \frac{1}{ab(a+b+c)} +\frac{1}{bc(a+b+c)}+\frac{1}{ca(a+b+c)}=\frac{1}{abc}.$$

31. Baltic Way 2004 $a,b,c>0,abc = 1,n\in\mathbb{N}$
$$\frac{1}{a^n+b^n+1}+\frac{1}{b^n+c^n+1}+\frac{1}{c^n+a^n+1}\leq 1$$

Solution :

Let $x=a^{n/3},y=b^{n/3},z=c^{n/3}.$ Then $xyz=1.$ Thus the inequality is equivalent to # 87.

92. IMO Short List 1996 (a,b,c>0,abc=1)
$$\frac{ab}{a^5+b^5+ab}+\frac{bc}{b^5+c^5+bc}+\frac{ca}{c^5+a^5+ca}\leq 1$$

Solution :

We can easily show that
$$ab(a^3+b^3)\leq a^5+b^5.$$
Thus
$$\frac{ab}{a^5+b^5+ab}\leq \frac{1}{a^3+b^3+abc}. $$
Hence the inequality is a direct consequence of #87.

ได้ข้อ 7 แล้วครับ

7. Romania 2005 $a,b,c>0, \,abc\geq 1$
$$\frac{1}{a+b+1}+\frac{1}{b+c+1}+\frac{1}{c+a+1}\leq 1$$

Solution :

Let $k=\sqrt[3]{abc}$ and $\displaystyle{ x=\frac{a}{k},y=\frac{b}{k},z=\frac{c}{k}}$. Then $k\geq 1$ and $xyz=1$. Thus

$\displaystyle{ LHS = \frac{1}{kx+ky+1}+\frac{1}{ky+kz+1}+\frac{1}{kz+kx+1}}$

$\displaystyle{ \leq \frac{1}{x+y+1}+\frac{1}{y+z+1}+\frac{1}{z+x+1}}$

$\leq 1.$

Note that the last inequality is just a special case of # 31.