Genelleştirilmiş USAMO 2017 #6

[Hüseyin Yiğit EMEKÇİ]

(Hüseyin Emekçi)
$a,b,c,d,n,k$ negatif olmayan reel sayılar, $n>0$ ve $a+b+c+d=n^k$ olmak üzere

$$\dfrac{a}{\dfrac{kb^{k+1}}{n}+n^k}+\dfrac{b}{\dfrac{kc^{k+1}}{n}+n^k}+\dfrac{c}{\dfrac{kd^{k+1}}{n}+n^k}+\dfrac{d}{\dfrac{ka^{k+1}}{n}+n^k}\geq \dfrac{k(4-n^{k-1})+4}{4k+4}$$

olduğunu gösteriniz.

$a,b,c,d,k$ negatif olmayan reel sayılar olmak üzere $a+b+c+d=2^k$ ise

$$\dfrac{a}{\dfrac{kb^{k+1}}{2}+2^k}+\dfrac{b}{\dfrac{kc^{k+1}}{2}+2^k}+\dfrac{c}{\dfrac{kd^{k+1}}{2}+2^k}+\dfrac{d}{\dfrac{ka^{k+1}}{2}+2^k}\geq \dfrac{k(1-2^{k-3})+1}{k+1}$$

olduğunu gösteriniz.

$a,b,c,d,k$ negatif olmayan reel sayılar olmak üzere $a+b+c+d=k^k$ ise

$$\dfrac{a}{b^{k+1}+k^k}+\dfrac{b}{c^{k+1}+k^k}+\dfrac{c}{d^{k+1}+k^k}+\dfrac{d}{a^{k+1}+k^k}\geq \dfrac{k(4-k^{k-1})+4}{4k+4}$$

olduğunu gösteriniz.

[Hüseyin Yiğit EMEKÇİ]

(Hüseyin Emekçi)
$$\dfrac{a}{\dfrac{kb^{k+1}}{n}+n^k}+\dfrac{b}{\dfrac{kc^{k+1}}{n}+n^k}+\dfrac{c}{\dfrac{kd^{k+1}}{n}+n^k}+\dfrac{d}{\dfrac{ka^{k+1}}{n}+n^k}\geq p$$

$$\dfrac{a}{\dfrac{kb^{k+1}}{n}+n^k}-\dfrac{a}{n^k}+\dfrac{b}{\dfrac{kc^{k+1}}{n}+n^k}-\dfrac{b}{n^k}+\dfrac{c}{\dfrac{kd^{k+1}}{n}+n^k}-\dfrac{c}{n^k}+\dfrac{d}{\dfrac{ka^{k+1}}{n}+n^k}-\dfrac{d}{n^k}\geq p-\dfrac{a+b+c+d}{n^k}=p-1$$

$$=a\left(\dfrac{1}{\dfrac{kb^{k+1}}{n}+n^k}-\dfrac{1}{n^k}\right)+b\left(\dfrac{1}{\dfrac{kc^{k+1}}{n}+n^k}-\dfrac{1}{n^k}\right)+c\left(\dfrac{1}{\dfrac{kd^{k+1}}{n}+n^k}-\dfrac{1}{n^k}\right)+d\left(\dfrac{1}{\dfrac{ka^{k+1}}{n}+n^k}-\dfrac{1}{n^k}\right)$$

$$=-\left(\dfrac{akb^{k+1}}{n^{k+1}\left(n^k+\dfrac{kb^{k+1}}{n}\right)}+\dfrac{bkc^{k+1}}{n^{k+1}\left(n^k+\dfrac{kc^{k+1}}{n}\right)}+\dfrac{ckd^{k+1}}{n^{k+1}\left(n^k+\dfrac{kd^{k+1}}{n}\right)}+\dfrac{dka^{k+1}}{n^{k+1}\left(n^k+\dfrac{ka^{k+1}}{n}\right)}\right)\geq p-1$$

$$\sum_{cyc}{\left(\dfrac{akb^{k+1}}{n^{k+1}\left(n^k+\dfrac{kb^{k+1}}{n}\right)}\right)}\leq 1-p$$

$$\sum_{cyc}{\left(\dfrac{akb^{k+1}}{n^{k+1}\left(n^k+\dfrac{kb^{k+1}}{n}\right)}\right)}=\sum_{cyc}{\left(\dfrac{akb^{k+1}}{n^{k+1}\left(n^k+\underbrace{\dfrac{b^{k+1}}{n}+\dfrac{b^{k+1}}{n}+\cdots+\dfrac{b^{k+1}}{n}}_{k}\right)}\right)}\overbrace{\leq}^{AM-GM}\sum_{cyc}{\left(\dfrac{akb^{k+1}}{n^{k+1}(k+1)b^k}\right)}$$

$$=\sum_{cyc}{\dfrac{akb}{n^{k+1}(k+1)}}=\dfrac{k}{n^{k+1}(k+1)}(ab+bc+ca+ad))=\dfrac{k}{n^{k+1}(k+1)}(a+c)(b+d)\overbrace{\leq}^{AM-GM}\dfrac{k\left(\dfrac{a+b+c+d}{2}\right)^2}{n^{k+1}(k+1)}$$

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

$$\rightarrow p\leq \dfrac{4k+4-kn^{k-1}}{4k+4}=\dfrac{k(4-n^{k-1})+4}{4k+4}$$

İspat biter.

[Hüseyin Yiğit EMEKÇİ]

Genelleştirme 4
$a_{1},a_{2},\cdots,a_{p},n,k$ negatif olmayan reel sayılar, $n>0, p\geq 3$ ve $\sum_{cyc}{a_{1}}=n^k$ olmak üzere

$$\sum_{cyc}{\dfrac{a_{1}}{\dfrac{ka_{2}^{k+1}}{n}+n^k}}\geq  \dfrac{k(p-n^{k-1})+p}{p(k+1)}$$

olduğunu gösteriniz.