From AM-GM inequality we have $a^2+1 \ge 2a$, thus: $$ \frac{ab}{a^2-\frac{4}{3}a+\frac{4}{3}} \le \frac{3ab}{2a+1} $$The main trick is the following: $$ \frac{3ab}{2a+1} = \frac{1}{2}(3b-\frac{3b}{2a+1}) $$Using it we just need to prove the following: \begin{align*} \frac{1}{2}( 3(a+b+c+d) -(\frac{3b}{2a+1}+\frac{3c}{2b+1}+\frac{3d}{2c+1}+\frac{3a}{2d+1})) \le 4 \\ \frac{3b}{2a+1}+\frac{3c}{2b+1}+\frac{3d}{2c+1}+\frac{3a}{2d+1} \ge 4 \\ \frac{b}{2a+1}+\frac{c}{2b+1}+\frac{d}{2c+1}+\frac{a}{2d+1} \ge \frac{4}{3} \\ \end{align*}On the other hand from Titu's lemma, we can conclude that: $$ \frac{b^2}{2ab+b}+\frac{c^2}{2bc+c}+\frac{d^2}{2cd+d}+\frac{a^2}{2ad+a} \ge \frac{(a+b+c+d)^2}{4+2(a+c)(b+d)} $$This means we just need to show: \begin{align*} \frac{16}{4+2(a+c)(b+d)} \ge \frac{4}{3} \\ 12 \ge 4+2(a+c)(b+d) \\ 4 \ge (a+c)(b+d) \\ 16 \ge 4(a+c)(b+d) \\ (a+b+c+d)^2 \ge 4(a+c)(b+d) \end{align*}Setting $a+c=x$, $b+d=y$, we get $(x+y)^2 \ge 4xy$, which is obvious.

matinyousefi wrote: Positive real numbers $a, b, c$ and $d$ are given such that $a+b+c+d = 4$ prove that $$\frac{ab}{a^2-\frac{4}{3}a+\frac{4}{3}} + \frac{bc}{b^2-\frac{4}{3}b+ \frac{4}{3}} + \frac{cd}{c^2-\frac{4}{3}c+ \frac{4}{3}} + \frac{da}{d^2-\frac{4}{3}d+ \frac{4}{3}}\leq 4.$$

Positive real numbers $a, b, c$ are given such that $a+b+c =3$ prove that $$ \frac{ab}{a^2-\frac{2}{3}a+\frac{2}{3}}+\frac{bc}{b^2-\frac{2}{3}b+\frac{2}{3}} +\frac{ca}{c^2-\frac{2}{3}c+ \frac{2}{3}} \leq 3 $$$$\frac{bc}{a^2-\frac{4}{3}a+\frac{4}{3}}+\frac{ca}{b^2-\frac{4}{3}b+\frac{4}{3}} +\frac{ab}{c^2-\frac{4}{3}c+ \frac{4}{3}} \leq 3 $$Positive real numbers $a, b, c$ are given such that $a^2+b^2+c^2 =3$ prove that $$ \frac{ab}{a^2-a+1}+\frac{bc}{b^2-b+1} +\frac{ca}{c^2-c+1} \leq a+b+c $$$$ \frac{ab}{a^2-\frac{2}{3}a+\frac{2}{3}}+\frac{bc}{b^2-\frac{2}{3}b+\frac{2}{3}} +\frac{ca}{c^2-\frac{2}{3}c+ \frac{2}{3}} \leq a+b+c $$$$\frac{bc}{a^2-\frac{4}{3}a+\frac{4}{3}}+\frac{ca}{b^2-\frac{4}{3}b+\frac{4}{3}} +\frac{ab}{c^2-\frac{4}{3}c+ \frac{4}{3}} \leq a+b+c $$

matinyousefi wrote: Positive real numbers $a, b, c$ and $d$ are given such that $a+b+c+d = 4$ prove that $$\frac{ab}{a^2-\frac{4}{3}a+\frac{4}{3}} + \frac{bc}{b^2-\frac{4}{3}b+ \frac{4}{3}} + \frac{cd}{c^2-\frac{4}{3}c+ \frac{4}{3}} + \frac{da}{d^2-\frac{4}{3}d+ \frac{4}{3}}\leq 4.$$ By tangent-line on the function $f(x)=\frac x{x^2-\frac43x+\frac43}$ we have that: $$f(x)-\frac13x-\frac23=-\frac{(x-1)^2(3x+8)}{9\left(x-\frac23\right)^2+8}\le0.$$Thus: $$bf(a)+cf(b)+df(c)+af(d)\le\frac13(ab+bc+cd+da)+\frac23(a+b+c+d)=\frac13(a+c)(b+d)+\frac83\le\frac13\left(\frac{a+b+c+d}2\right)^2+\frac83=4.$$

Positive real numbers $a, b, c$ are given such that $a^2+b^2+c^2 =3$ prove that$$ \frac{ab}{a^2-\frac{4}{3}a+\frac{4}{3}}+\frac{bc}{b^2-\frac{4}{3}b+\frac{4}{3}} +\frac{ca}{c^2-\frac{4}{3}c+ \frac{4}{3}} \leq a+b+c $$$$ \frac{ab}{a^2-\frac{5}{3}a+\frac{5}{3}}+\frac{bc}{b^2-\frac{5}{3}b+\frac{5}{3}} +\frac{ca}{c^2-\frac{5}{3}c+ \frac{5}{3}} \leq a+b+c $$

Note that $a^2 - \frac{4a}{3} + \frac{4}{3} \ge \frac{2a}{3} + \frac{1}{3}$ with AM-GM for $a^2,1$. So $\frac{ab}{a^2-\frac{4}{3}a+\frac{4}{3}} + \frac{bc}{b^2-\frac{4}{3}b+ \frac{4}{3}} + \frac{cd}{c^2-\frac{4}{3}c+ \frac{4}{3}} + \frac{da}{d^2-\frac{4}{3}d+ \frac{4}{3}} \le \frac{3ab}{2a+1} + \frac{3bc}{2b+1} + \frac{3cd}{2c+1} + \frac{3da}{2d+1}$. Note that $\frac{3ab}{2a+1} = \frac{3b-\frac{3b}{2a+1}}{2}$ so we need to prove $\frac{4}{3} \le \frac{b}{2a+1} + \frac{c}{2b+1} + \frac{d}{2c+1} + \frac{a}{2d+1}$. Now with titu we have $\frac{b}{2a+1} + \frac{c}{2b+1} + \frac{d}{2c+1} \ge \frac{(a+b+c+d)^2}{2ab+2bc+2cd+2da + a+b+c+d} = \frac{16}{2(ab+bc+cd+da) + 4}$ so we need to prove $4 \ge ab+bc+cd+da$ which is true since we have $(a+b+c+d)^2 \ge 4(ab+bc+cd+da)$. we're Done.

Old Idea We can see $a^2-\frac{4}{3}a+\frac{4}{3} \geq 2a-\frac{4}{3}a+\frac{1}{3} = \frac{2a+1}{3}$ ( ) With Placement ( ) in problem we can see : $$\sum \frac{ab}{a^2-\frac{4}{3}a+\frac{4}{3}} \leq \sum \frac{3ab}{2a+1} $$ Now we want prove $ \sum \frac{3ab}{2a+1} \leq 4$ . We can use this old trick ! $$ \sum \frac{3ab}{2a+1} - \frac{3a}{2} \leq 4-\frac{3(a+b+c+d)}{2}=6$$ Now we should prove $\sum \frac{b}{2a+1} \geq \frac{4}{3}$ . By C-S we can write : $$\sum \frac{b}{2a+1} \geq \frac{(a+b+c+d)^2}{\sum b(2a+1)} = \frac{8}{2+\sum ab} $$ Now we only need prove $\sum ab \leq 4$ . And it is very easy by AM-GM ! $\blacksquare$

The original problem is a special case of the following generalization where $$k=3,\lambda=1,\beta=1,p=4$$.

Generalization 1 Let $a, b, c,d$ be positive reals such that $a+b+c+d = p$. Then prove that $$\dfrac{\lambda ab}{a^2-\dfrac{k-\beta }{k}a+\dfrac{k+\lambda }{k}} +\dfrac{\lambda bc}{b^2-\dfrac{k-\beta }{k}b+\dfrac{k+\lambda }{k}}+ \dfrac{\lambda cd}{c^2-\dfrac{k-\beta }{k}c+\dfrac{k+\lambda }{k}}+\dfrac{\lambda da}{d^2-\dfrac{k-\beta }{k}d+\dfrac{k+\lambda }{k}} \leq \dfrac{\lambda p^2k}{\left(k-\beta\right)+4\lambda}$$

matinyousefi wrote: Positive real numbers $a, b, c$ and $d$ are given such that $a+b+c+d = 4$ prove that $$\frac{ab}{a^2-\frac{4}{3}a+\frac{4}{3}} + \frac{bc}{b^2-\frac{4}{3}b+ \frac{4}{3}} + \frac{cd}{c^2-\frac{4}{3}c+ \frac{4}{3}} + \frac{da}{d^2-\frac{4}{3}d+ \frac{4}{3}}\leq 4.$$ Proposed by Mohammad Sharifi Kiasari