Hope this is right...(idk if i used rearrangement correctly-this is the first time i actually used it in a problem)

Note: In the solution, when I said $LS$, I meant of the original inequality stated in the problem.

We have $$\frac{7+2b}{1+a}+\frac{7+2c}{1+b}+\frac{7+2a}{1+c}=2 \left (\frac{1+b}{1+a}+\frac{1+c}{1+b}+\frac{1+a}{1+c} \right) +5 \left(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c} \right) \geq 6\sqrt[3]{\frac{1+b}{1+a}+\frac{1+c}{1+b}+\frac{1+a}{1+c}}+\frac{5.9}{1+a+1+b+1+c}=\frac{69}{4}$$

tchebytchev wrote: We have $$\frac{7+2b}{1+a}+\frac{7+2c}{1+b}+\frac{7+2a}{1+c}=2 \left (\frac{1+b}{1+a}+\frac{1+c}{1+b}+\frac{1+a}{1+c} \right) +5 \left(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c} \right) \geq 6\sqrt[3]{\frac{1+b}{1+a}\cdot \frac{1+c}{1+b}\cdot \frac{1+a}{1+c}}+\frac{5.9}{1+a+1+b+1+c}=\frac{69}{4}$$

tchebytchev wrote: We have $$\frac{7+2b}{1+a}+\frac{7+2c}{1+b}+\frac{7+2a}{1+c}=2 \left (\frac{1+b}{1+a}+\frac{1+c}{1+b}+\frac{1+a}{1+c} \right) +5 \left(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c} \right) \geq 6\sqrt[3]{\frac{1+b}{1+a}+\frac{1+c}{1+b}+\frac{1+a}{1+c}}+\frac{5.9}{1+a+1+b+1+c}=\frac{69}{4}$$ My solution

Nice solution

My Solution. By C-S, $\sum \frac{1}{1+a}\ge \frac{9}{3+a+b+c}=9/4.$ Again by C-S, $\sum \frac{b^2}{b+ab} \ge \frac{1}{a+b+c+ab+bc+ca} \ge 3/4.$ Mutiplying with respective numerals and adding give the desired inequality.

too easy!!!! why was it choosen as a TST?

aditya21 wrote: too easy!!!! why was it choosen as a TST? You can see it is first question

With the conditions $a,b,c\in\mathbb{R^+}$ and $a+b+c=1$, prove that \[\left(\frac{7+2b}{1+a}\right)\left(\frac{7+2c}{1+b}\right)\left(\frac{7+2a}{1+c}\right)\geq\left(\frac{23}{4}\right)^3\]

$\frac{7+2b}{1+a}=\frac{2+2b}{1+a}+\frac{5}{1+a}=\frac{2(b+1)}{a+1}+\frac{5}{a+1}$ So, we have: with: $a+1=x;b+1=y;c+1=z=>\prod (\frac{2y}{x}+\frac{5}{x})\geqslant (\frac{23}{4})^3;x+y+z=4$

Thank phantranhuongth. With the conditions $a,b,c\geq 0$ and $a+b+c=1$, prove that \[\frac{7+2b}{1+a}+\frac{7+2c}{1+b}+\frac{7+2a}{1+c}\leq\frac{39}{2}\]

sqing wrote: Thank phantranhuongth. With the conditions $a,b,c\geq 0$ and $a+b+c=1$, prove that \[\frac{7+2b}{1+a}+\frac{7+2c}{1+b}+\frac{7+2a}{1+c}\leq\frac{39}{2}\] http://www.artofproblemsolving.com/community/c4h1061866p4781094：

By rearrangement and Cauchy-Schwarz, $$\frac{7+2b}{1+a}+\frac{7+2c}{1+b}+\frac{7+2a}{1+c} \ge \frac{7+2a}{1+a} + \frac{7+2b}{1+b} + \frac{7+2c}{1+c} = 6 + \frac{5}{1+a} + \frac{5}{1+b} + \frac{5}{1+c} \ge 6 + 5 \cdot \frac{9}{3+a+b+c} = 6 + \frac{45}{4} = \frac{69}{4}.$$

Is it just me, or does it seem like the numerator terms was specifically chosen so that the constant on the RHS would be equal to $69$?

IstekOlympiadTeam wrote: With the conditions $a,b,c\in\mathbb{R^+}$ and $a+b+c=1$, prove that \[\frac{7+2b}{1+a}+\frac{7+2c}{1+b}+\frac{7+2a}{1+c}\geq\frac{69}{4}\] By the CS inequality \[ ((7+2b)(1+a)+(7+2c)(1+b)+(7+2a)(1+c)) \left( \frac{7+2b}{1+a}+\frac{7+2c}{1+b}+\frac{7+2a}{1+c} \right) \geq (7+2b+7+2c+7+2a)^2 = 23^2, \] so \[ \left( \frac{7+2b}{1+a}+\frac{7+2c}{1+b}+\frac{7+2a}{1+c} \right) \geq \frac{23^2}{(7+2b)(1+a)+(7+2c)(1+b)+(7+2a)(1+c)}=\frac{23^2}{30+2ab+2bc+2ac}. \] It remains to be proven that $\frac{23^2}{30+2ab+2bc+2ac}\geq \frac{69}{4} \Leftrightarrow 30+2ab+2bc+2ac \leq \frac{92}{3} \Leftrightarrow ab+bc+ca \leq \frac{1}{3}$, which is trivial by $AM-GM$. Done.

IstekOlympiadTeam wrote: With the conditions $a,b,c\in\mathbb{R^+}$ and $a+b+c=1$, prove that \[\frac{7+2b}{1+a}+\frac{7+2c}{1+b}+\frac{7+2a}{1+c}\geq\frac{69}{4}\] By AM-GM, we have $\frac{1}{1+a}+\frac{9}{16}(1+a)\geq 2\sqrt{\frac{1}{1+a}\cdot \frac{9}{16}(1+a)} = \frac 32$. The equality is attained when $\frac{1}{1+a} =\frac{9}{16}(1+a)\ (a>0)\Longleftrightarrow a=\frac 13$, yielding $\frac{1}{1+a}\geq -\frac{9}{16}a+\frac{15}{16}.$ Thus we obtain $\frac{7+2b}{1+a}\geq \frac{3}{16}(5-3a)(7+2b)=\frac{3}{16}(35+10b-21a-6ab).$ $\therefore \frac{7+2b}{1+a}+\frac{7+2c}{1+b}+\frac{7+2a}{1+c}\geq \frac{3}{16}\{35\cdot 3+10(b+c+a)-21(a+b+c)-6(ab+bc+ca)\}$ $=\frac{3}{16}\{105-11(a+b+c)-6(ab+bc+ca)\}\geq\frac{3}{16}\{105-11(a+b+c)-2(a+b+c)^2\}$ $\because (a+b+c)^2\geq 3(ab+bc+ca)\Longleftrightarrow \frac{1}{2}\{(a-b)^2+(b-c)^2+(c-a)^2\}\geq 0.$ $\therefore \frac{7+2b}{1+a}+\frac{7+2c}{1+b}+\frac{7+2a}{1+c}\geq \frac{3}{16}(105-11\cdot 1-2\cdot1^2)=\frac{69}{4}.$ The equality is attained when $a=b=c=\frac 13.$

I also used rearrangement and Cauchy Engel. Nice and ez

My solution: $\sum_{cyc} \frac{7+2b}{1+a}=\sum_{cyc} \frac{7}{1+a} + \frac{2b}{1+a}$ Then $$\sum_{cyc} \frac {7}{1+a} \geq 7( \frac{9}{a+b+c+3})= \frac{63}{4}$$by Titu's Lemma. Also $$ \sum_{cyc} \frac{2b}{1+a}=2(\sum_{cyc} \frac {b}{1+a}) \geq 2(\frac{(a+b+c)^2}{ab+bc+ca+a+b+c}$$by Titu's Lemma. Since $$a+b+c=1=(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ac \geq 3(ab+bc+ca)$$by the rearrangement inequality. Therefore, $$ 2(\frac{(a+b+c)^2}{ab+bc+ca+a+b+c}) \geq \frac{3}{2}$$The result follows.

sqing wrote: Thank phantranhuongth. With the conditions $a,b,c\geq 0$ and $a+b+c=1$, prove that \[\frac{7+2b}{1+a}+\frac{7+2c}{1+b}+\frac{7+2a}{1+c}\leq\frac{39}{2}\] https://artofproblemsolving.com/community/c6h1501761p12611725: Let $a, b, c>0, a+b+c=1$. Prove that \[\frac{7+2a}{1+c}+\frac{7+2b}{1+a}+\frac{7+2c}{1+b}\le\frac{23}{4(ab+bc+ca)}\]

IstekOlympiadTeam wrote: With the conditions $a,b,c\in\mathbb{R^+}$ and $a+b+c=1$, prove that \[\frac{7+2b}{1+a}+\frac{7+2c}{1+b}+\frac{7+2a}{1+c}\geq\frac{69}{4}\] Direct Chebyshev and AM-HM $LHS\geq \frac13(7+2a+7+2b+7+2c)\left(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}\right)\geq\frac{23}{3}\left(\frac{9}{3+a+b+c}\right)=\frac{69}{4}$

Rewrite is as $(7+2b)^2/((a+1)(7+2b))+(7+2c)^2/((b+1)(7+2c))+(7+2a)^2/((c+1)(7+2a))=>69/4$. Apply Cauchy Schwarz in Titu form to get $LHS=>23^2/(30+2(ab+bc+ac))$ (we used that $a+b+c=1$). Now it's equivalent to prove that $1=>3(ab+bc+ac)$. But if we plug $1=1^2=(a+b+c)^2$, we have to prove that $a^2+b^2+c^2-ab-bc-ac=>0$, which is trivial.

All the solutions you guys propose are beautiful and more or less equal! I found really helpful to set a+1=x, b+1=y and c+1=z. After that it is much easier to get the required

Not bad ! Let $\frac{7+2b}{1+a}+\frac{7+2c}{1+b}+\frac{7+2a}{1+c} = A$ By C-S we can write : $$(\sum \frac{7+2b}{1+a})(\sum (1+a)(7+2b)) \geq (\sum(7+2b))^2$$ Or we need prove : $A \geq \frac{23^2}{9+2\sum ab} \geq \frac{69}{4}$ If we write all of the phrase we should prove : $$92 \geq 90 + 6\sum ab$$or $\frac{1}{3} \geq ab+bc+ca$ We know $(a+b+c)^2= 1 \geq 3(ab+bc+ca)$ or $\frac{1}{3} \geq ab+bc+ca$ $\blacksquare$

the old days are beautiful than nowdays $$\frac{7+2b}{1+a}+\frac{7+2c}{1+b}+\frac{7+2a}{1+c}\geq\frac{69}{4}$$$$C-S$$$$\sum{\frac{7}{1+a}}\geq \frac{63}{4}$$We need to prove the other side for 3/2 $$C-S$$$$\sum{\frac{2b}{1+a}}\geq \frac{2}{1+ab+bc+ca}$$$$\frac{2}{1+ab+bc+ca} \geq \frac{3}{2}$$finally we get $$\frac{1}{3} \geq ab+bc+ca$$and it is true because $$\frac{(a+b+c)^2}{3} \geq ab+bc+ca$$

Nice problem

I think they give this question as a gift for a students ok Let's sayt what will I do I will make suitale it for Titu's lemma from first observation... \[ \sum_{\text{cyc}} \frac{7 + 2b}{1 + a} = 5 \sum_{\text{cyc}} \frac{1}{1 + a} + 2 \sum_{\text{cyc}} \frac{1 + b}{1 + a}. \] For the first term, \( 5 \sum_{\text{cyc}} \frac{1}{1 + a} \), we apply Titu’s Lemma: \[ \sum_{\text{cyc}} \frac{1}{1 + a} \geq \frac{(1 + 1 + 1)^2}{(1 + a) + (1 + b) + (1 + c)} = \frac{9}{a + b + c + 3}. \]Thus, \[ 5 \sum_{\text{cyc}} \frac{1}{1 + a} \geq 5 \cdot \frac{9}{a + b + c + 3} = \frac{45}{a + b + c + 3}. \]Since \( a + b + c = 1 \), this simplifies to: \[ \frac{45}{a + b + c + 3} = \frac{45}{4}. \] For the second term, \( 2 \sum_{\text{cyc}} \frac{1 + b}{1 + a} \), we apply AM-GM: \[ \sum_{\text{cyc}} \frac{1 + b}{1 + a} \geq 3 \sqrt[3]{\frac{(1 + b)(1 + c)(1 + a)}{(1 + a)(1 + b)(1 + c)}} = 3. \]Thus: \[ 2 \sum_{\text{cyc}} \frac{1 + b}{1 + a} \geq 2 \cdot 3 = 6. \] Adding these results: \[ \text{LHS} \geq \frac{45}{4} + 6 = \frac{69}{4}. \] The inequality holds, and we are done.

KartikPatekar wrote: My Solution. By C-S, $\sum \frac{1}{1+a}\ge \frac{9}{3+a+b+c}=9/4.$ Again by C-S, $\sum \frac{b^2}{b+ab} \ge \frac{1}{a+b+c+ab+bc+ca} \ge 3/4.$ Mutiplying with respective numerals and adding give the desired inequality. Doesn’t make sense cuz I’m not good at inequalities. Can someone explain