Let $a,b,c,d \in [0,1]$. Prove $$ \frac {a}{1+b}+\frac{b}{1+c}+\frac{c}{1+d}+\frac{d}{1+a}+abcd\le3$$
Hi. Im a school student so ive been unable to look for ways to apply this problem. I attempted to use AM-GN on 1+a etc but the bound $2\sqrt{a}$ was too large and was easily larger than 3. Thus id like some help solving this problem.
On
We need to prove that $$\sum_{cyc}\left(\frac{a}{1+b}-a\right)+abcd\leq3-a-b-c-d$$ or $$3-a-b-c-d+\sum_{cyc}\frac{ab}{1+b}\geq abcd.$$ Now, by C-S and the given condition $$\sum_{cyc}\frac{ab}{1+b}=\sum_{cyc}\frac{abcd}{cd+bcd}\geq\frac{16abcd}{\sum\limits_{cyc}(cd+bcd)}\geq\frac{16abcd}{8}=2abcd.$$ Thus, it's enough to prove that $$3-a-b-c-d+abcd\geq0,$$ which is a linear inequality of all variables.
Thus, $$\min_{\{a,b,c,d\}\subset[0,1]}(3-a-b-c-d+abcd)=\min_{\{a,b,c,d\}\subset\{0,1\}}(3-a-b-c-d+abcd)=0$$ and we are done!
For $a,b,c,d \in [0,1]$, let $$ f(a,b,c,d) = \frac{a}{1+b} + \frac{b}{1+c} + \frac{c}{1+d} + \frac{d}{1+a} + abcd $$ The goal is to prove $f(a,b,c,d) \le 3$, for all $a,b,c,d \in [0,1]$.
For fixed $b,c,d \in [0,1]$, let $g(a) = f(a,b,c,d)$.
Then $$g''(a) = \frac{2d}{(a+1)^3}$$ hence $g''(a) \ge 0$, for all $a \in [0,1]$.
It follows that $g$ is convex on $[0,1]$, hence the maximum value of $g$ on $[0,1]$ is achieved at one of the endpoints.
Thus, for fixed $b,c,d$, the function $f(a,b,c,d)$ is maximized at $a=0$ or $a=1$.
The same reasoning applies to each of the variables $b,c,d$, with the other $3$ variables held fixed.
Thus, to prove $f(a,b,c,d) \le 3$, for all $a,b,c,d \in [0,1]$, it suffices to verify that $f(a,b,c,d) \le 3$, for all $a,b,c,d \in \{0,1\}$.
If any of $a,b,c,d$ is zero, the expression $$ \frac{a}{1+b} + \frac{b}{1+c} + \frac{c}{1+d} + \frac{d}{1+a} + abcd $$ reduces to a sum of $3$ fractions, each of which is at most $1$, hence the sum is at most $3$.
If $a,b,c,d$ are all equal to $1$, the expression $$ \frac{a}{1+b} + \frac{b}{1+c} + \frac{c}{1+d} + \frac{d}{1+a} + abcd $$ evaluates to exactly $3$.
Thus, the claimed inequality is proved.