Genelleştirilmiş USAMO 2003 #5 {çözüldü}

[Hüseyin Yiğit EMEKÇİ]

$a,b,c$ pozitif reeller olmak üzere

$$\sum_{cyc}{\dfrac{\left((k+1)a+kb+kc\right)^2}{(k+1)a^2+k(b+c)^2}}\leq 6k+\dfrac{3}{2k+1}+1$$

olduğunu gösteriniz.

[Hüseyin Yiğit EMEKÇİ]

Eşitsizliğin homojen özellikte olduğuna dikkat edelim. O zaman $a+b+c=1$ olsun.

$$\sum_{cyc}{\dfrac{\left((k+1)a+kb+kc\right)^2}{(k+1)a^2+k(b+c)^2}}=\sum_{cyc}{\dfrac{(a+k)^2}{(k+1)a^2+k(1-a)^2}}=\sum_{cyc}{\dfrac{(a+k)^2}{(2k+1)a^2-2ak+k}}$$
$$=\sum_{cyc}{\left(\dfrac{1}{2k+1}+\dfrac{2ak+\dfrac{2ak}{2k+1}+k^2-\dfrac{k}{2k+1}}{(2k+1)a^2-2ak+k}\right)}=S$$

Paydadaki ifadeye bir eşitsizlik hamlesinde bulunalım.
$$(2k+1)a^2-2ak+k=(2k+1)a^2-2ak+\dfrac{k^2}{2k+1}+k-\dfrac{k^2}{2k+1}\overbrace{\geq}^{AGO} 2ak-2ak+k-\dfrac{k^2}{2k+1}=k-\dfrac{k^2}{2k+1}$$

$$S\leq \sum_{cyc}{\left(\dfrac{1}{2k+1}+\dfrac{2ak+\dfrac{2ak}{2k+1}+k^2-\dfrac{k}{2k+1}}{k-\dfrac{k^2}{2k+1}}\right)}=\dfrac{3}{2k+1}+\dfrac{2k(a+b+c)+\dfrac{2k}{2k+1}(a+b+c)+3k^2-\dfrac{3k}{2k+1}}{k-\dfrac{k^2}{2k+1}}$$
$$=\dfrac{3}{2k+1}+\dfrac{2k+\dfrac{2k}{2k+1}+3k^2-\dfrac{3k}{2k+1}}{k-\dfrac{k^2}{2k+1}}=\dfrac{3}{2k+1}+\dfrac{(2k+1)\left(3k^2+2k-\dfrac{k}{2k+1}\right)}{k^2+k}$$
$$=\dfrac{3}{2k+1}+\dfrac{(2k+1)k(3k+2-\dfrac{1}{2k+1}}{k(k+1)}=\dfrac{3}{2k+1}+\dfrac{(2k+1)(3k+2-\dfrac{1}{2k+1})}{k+1}=\dfrac{3}{2k+1}+\dfrac{(2k+1)\left(\dfrac{6k^2+7k+1}{2k+1}\right)}{k+1}$$
$$=\dfrac{3}{2k+1}+\dfrac{6k^2+7k+1}{k+1}=\dfrac{3}{2k+1}+6k+1$$

[Hüseyin Yiğit EMEKÇİ]

Genelleştirme 2
$a_{1},a_{2},\cdots,a_{n}$ pozitif reeller olmak üzere


$$\sum_{cyc}{\dfrac{\left((k+1)a_{1}+k(a_{2}+a_{3}+\cdots+a_{n}\right)^2}{(k+1)a_{1}^2+k(a_{2}+a_{3}+\cdots+a_{n})}}\geq n(2k-1)+\dfrac{n}{2k+1}+4$$

olduğunu gösteriniz.

[Hüseyin Yiğit EMEKÇİ]

Genel problemin çözümü yine homojeniteden dolayı $\sum_{cyc}{a_{1}}=1$ atamasıyla başlıyor. Cebirsel manipülasyonlar sonucu


$$\leq \sum_{cyc}{\left(\dfrac{1}{2k+1}+\dfrac{2a_{1}k+\dfrac{2a_{1}k}{2k+1}+k^2-\dfrac{k}{2k+1}}{k-\dfrac{k^2}{2k+1}}\right)}=\dfrac{n}{2k+1}+\dfrac{2k(\sum_{cyc}{a_{1}})+\dfrac{2k}{2k+1}(\sum_{cyc}{a_{1}})+nk^2-\dfrac{nk}{2k+1}}{k-\dfrac{k^2}{2k+1}}$$


Sondaki ifadenin ispatı orijinal sorunun çözümînde yapılmıştır.

[Hüseyin Yiğit EMEKÇİ]

$$\dfrac{n}{2k+1}+\dfrac{2k+\dfrac{2k}{2k+1}+nk^2-\dfrac{nk}{2k+1}}{k-\dfrac{k^2}{2k+1}}=\dfrac{n}{2k+1}+\dfrac{\dfrac{k(2-n)}{2k+1}+k(2+nk)}{k-\dfrac{k^2}{2k+1}}$$
$$=\dfrac{n}{2k+1}+\dfrac{\dfrac{2-n}{2k+1}+2+nk}{\dfrac{k+1}{2k+1}}=\dfrac{n}{2k+1}+\dfrac{2nk^2+nk-n+4k+4}{k+1}$$
$$=\dfrac{n}{2k+1}+\dfrac{n(2k^2+k-1)}{k+1}+4=\dfrac{n}{2k+1}+\dfrac{n(2k-1)(k+1)}{k+1}+4$$
$$=\dfrac{n}{2k+1}+n(2k-1)+\dfrac{n}{2k+1}+4$$

olduğundan dolayı eşitsizlik çalışır.