Školjka

Neka je $A=a_1+a_2+a_3+a_4$, pa koristeći $AG$ nejednakost imamo $$A=a_1+a_2+a_3+a_4\ge 4\sqrt[4]{ a_1a_2a_3a_4}=4$$ Iz toga slijedi \begin{align*} A&\ge4 \\ 16A&\ge 15A+4\\ 16A+16&\ge 15A+20\\ \frac{16A+16}{3A+4}&\ge5\\ (A+1)\big(\frac{16}{3A+4}\big)&\ge5 \end{align*} Nadalje, koristeći $AH$ nejednakost imamo $$\frac{16}{3A+4}=\frac{1}{4A-A+4}=\frac{16}{(A-a_1+1)+(A-a_2+1)+(A-a_3+1)+(A-a_4+1)}\stackrel{\text{AH}}{\le}\frac{1}{A-a_1+1} + \frac{1}{A-a_2+1} + \frac{1}{A-a_3+1} + \frac{1}{A-a_4+1}$$ Pa imamo \begin{align*} 5\le(A+1)\big(\frac{16}{3A+4}\big)&\le(A+1)\bigg(\frac{1}{A-a_1+1} + \frac{1}{A-a_2+1} + \frac{1}{A-a_3+1} + \frac{1}{A-a_4+1}\bigg)\\ &=\frac{A+1}{A-a_1+1} + \frac{A+1}{A-a_2+1} + \frac{A+1}{A-a_3+1} + \frac{A+1}{A-a_4+1}\\ &=\frac{a_1+(a_2+a_3+a_4+1)}{a_2 + a_3 + a_4 + 1} + \frac{a_2+(a_1+a_3+a_4+1)}{a_1 + a_3 + a_4 + 1} + \frac{a_3+(a_1+a_2+a_4+1)}{a_1 + a_2 + a_4 + 1} + \frac{a_4+(a_1+a_2+a_3+1)}{a_1 + a_2 + a_3 + 1}\\ &=\bigg(\frac{a_1}{a_2 + a_3 + a_4 + 1}+1\bigg) + \bigg(\frac{a_2}{a_1 + a_3 + a_4 + 1}+1\bigg) + \bigg(\frac{a_3}{a_1 + a_2 + a_4 + 1} +1\bigg)+ \bigg(\frac{a_4}{a_1 + a_2 + a_3 + 1}+1\bigg)\\ &=\frac{a_1}{a_2 + a_3 + a_4 + 1} + \frac{a_2}{a_1 + a_3 + a_4 + 1} + \frac{a_3}{a_1 + a_2 + a_4 + 1} + \frac{a_4}{a_1 + a_2 + a_3 + 1}+4 \end{align*} Iz čega direktno slijedi tražena nejednakost; \begin{align*} &\frac{a_1}{a_2 + a_3 + a_4 + 1} + \frac{a_2}{a_1 + a_3 + a_4 + 1} + \frac{a_3}{a_1 + a_2 + a_4 + 1} + \frac{a_4}{a_1 + a_2 + a_3 + 1}+4\ge5\\ \Longleftrightarrow &\frac{a_1}{a_2 + a_3 + a_4 + 1} + \frac{a_2}{a_1 + a_3 + a_4 + 1} + \frac{a_3}{a_1 + a_2 + a_4 + 1} + \frac{a_4}{a_1 + a_2 + a_3 + 1}\ge 1 \end{align*} Čime smo dokazali traženu nejednakost i to je to. Najsss!