Let $a, b, c, d$ be non-negative real numbers such that \[\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}+\frac{1}{d+1}=3.\]Prove that \[3(ab+bc+ca+ad+bd+cd)+\frac{4}{a+b+c+d}\leqslant 5.\]Vasile Cîrtoaje and Leonard Giugiuc
Problem
Source: BMO Shortlist 2022, A3
Tags: algebra, inequalities
mihaig
14.05.2023 07:57
Thanks, Vlad! Authors: Prof. V. Cirtoaje and yours truly. EDIT: Must replace $3\sum_{\text{cyc}}ab$ by $$3(ab+bc+ca+ad+bd+cd).$$
mihaig
14.05.2023 08:25
Let $a, b, c, d$ be non-negative real numbers such that \[\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}+\frac{1}{d+1}=3.\]Prove that \[3(ab+bc+ca+ad+bd+cd)+\frac{4}{a+b+c+d}\leqslant 5.\]
Orestis_Lignos
14.05.2023 09:30
In what follows, all sums are supposed to be cyclic. The given condition rewrites as $3abcd+2\sum abc+\sum ab=1$. Claim: $\dfrac{2}{3} \leq \sum ab \leq 1$. Proof: The right inequality is evident since $a,b,c,d \geq 0$. For the left one, note that $abcd \leq \dfrac{(\displaystyle \sum ab)^2}{36},$ due to AM-GM inequality. Moreover, if we let $s_1=\dfrac{\displaystyle \sum a}{4},$ $s_2=\dfrac{\displaystyle \sum ab}{6}$ and $s_3=\dfrac{\displaystyle \sum abc}{4},$ then by Newton's inequality $s_3s_1 \leq s_2^2,$ which rewrites as $9(a+b+c+d)\sum abc \leq 4(\sum ab)^2$. However, $2\sum ab=(\sum a)^2-\sum a^2 \leq (\sum a)^2 - \dfrac{(\displaystyle \sum a)^2}{4}=\dfrac{3(\displaystyle \sum a)^2}{4},$ and so $\sum ab \leq \dfrac{3(\displaystyle \sum a)^2}{8}$. Therefore, $4(\sum ab)^2 \geq 9(a+b+c+d)\sum abc \geq 9\sqrt{\dfrac{8}{3}}\sqrt{\sum ab} \sum abc,$ which gives us that $\sum abc \leq \sqrt{\dfrac{2(\displaystyle \sum ab)^3}{27}}$. Therefore, if we let $\sum ab=k,$ $1=\sum ab+2 \sum abc+3abcd \leq f(k),$ where $f$ is a strictly icnreasing function such that $f(\dfrac{2}{3})=1,$ and so we have $f(k) \geq 1=f(\dfrac{2}{3}),$ which in turn implies $k \geq \dfrac{2}{3},$ as desired $\blacksquare$ To the problem, note that the given condition rewrites as $\sum \dfrac{a}{a+1}=1,$ and so by Cauchy-Schwarz, $1 =\sum \dfrac{a^2}{a^2+a} \geq \dfrac{(\displaystyle \sum a)^2}{\displaystyle \sum a^2+\sum a},$ which rewrites as $2k=2\sum ab \leq \sum a$. Thus, to prove the desired inequality it suffices to prove that $3k+\dfrac{2}{k} \leq 5,$ that is $(k-1)(3k-2) \leq 0,$ or equivalently $\dfrac{2}{3} \leq k \leq 1,$ which by our Claim is true, and so we are done.
mihaig
14.05.2023 09:43
Now let's take it seriously https://artofproblemsolving.com/community/c6t243f6h3071847_a_refinement_for_bmo_shortlist_2022_a3
oVlad
14.05.2023 12:16
mihaig wrote: Thanks, Vlad! Authors: Prof. V. Cirtoaje and yours truly. As soon as I saw it I knew you were an author! Congrats!
VicKmath7
14.05.2023 17:48
All sums we use below are symmetric.
Firstly, we focus on the condition; we can rewrite it as $1=\sum \frac{a} {a+1}=\sum \frac{a^2}{a^2+a} \geq \frac{(a+b+c+d)^2}{a^2+b^2+c^2+d^2+a+b+c+d}$, which yields that $a+b+c+d \geq 2(ab+bc+cd+da+ac+bd):=2s$. Hence, we have to prove $3s+\frac{4}{2s}\leq 5 \iff (3s-2)(s-1) \leq 0 \iff s \in [\frac{2}{3};1]$.
The upper bound follows directly from the given condition by clearing the denominators and expanding the brackets, so we focus on the second one. Notice that since $\sum \frac{a}{a+1}=1$, we can find $x, y, z, t>0$, such that $\frac{a} {a+1}=\frac{x}{x+y+z+t}$, i.e. $a=\frac{x} {y+z+t}$ and similarly for $b, c, d$. Now we have to prove that $\sum \frac{xy} {(y+z+t)(x+z+t)} \geq \frac{2}{3}$. Clearing the denominators and expanding gives that we want $\sum x^3y+2\sum x^2y^2 \geq 2 \sum xyz^2$, which follows by Muirhead since $(3,1,0,0)$ and $(2,2,0,0)$ majorize $(2,1,1,0)$, so we are done.
mihaig
15.05.2023 05:00
oVlad wrote: mihaig wrote: Thanks, Vlad! Authors: Prof. V. Cirtoaje and yours truly. As soon as I saw it I knew you were an author! Congrats! Thank you, Vlad!
mihaig
15.05.2023 05:24
Orestis_Lignos wrote: In what follows, all sums are supposed to be cyclic. The given condition rewrites as $3abcd+2\sum abc+\sum ab=1$. Claim: $\dfrac{2}{3} \leq \sum ab \leq 1$. Proof: The right inequality is evident since $a,b,c,d \geq 0$. For the left one, note that $abcd \leq \dfrac{(\displaystyle \sum ab)^2}{36},$ due to AM-GM inequality. Moreover, if we let $s_1=\dfrac{\displaystyle \sum a}{4},$ $s_2=\dfrac{\displaystyle \sum ab}{6}$ and $s_3=\dfrac{\displaystyle \sum abc}{4},$ then by Newton's inequality $s_3s_1 \leq s_2^2,$ which rewrites as $9(a+b+c+d)\sum abc \leq 4(\sum ab)^2$. However, $2\sum ab=(\sum a)^2-\sum a^2 \leq (\sum a)^2 - \dfrac{(\displaystyle \sum a)^2}{4}=\dfrac{3(\displaystyle \sum a)^2}{4},$ and so $\sum ab \leq \dfrac{3(\displaystyle \sum a)^2}{8}$. Therefore, $4(\sum ab)^2 \geq 9(a+b+c+d)\sum abc \geq 9\sqrt{\dfrac{8}{3}}\sqrt{\sum ab} \sum abc,$ which gives us that $\sum abc \leq \sqrt{\dfrac{2(\displaystyle \sum ab)^3}{27}}$. Therefore, if we let $\sum ab=k,$ $1=\sum ab+2 \sum abc+3abcd \leq f(k),$ where $f$ is a strictly icnreasing function such that $f(\dfrac{2}{3})=1,$ and so we have $f(k) \geq 1=f(\dfrac{2}{3}),$ which in turn implies $k \geq \dfrac{2}{3},$ as desired $\blacksquare$ To the problem, note that the given condition rewrites as $\sum \dfrac{a}{a+1}=1,$ and so by Cauchy-Schwarz, $1 =\sum \dfrac{a^2}{a^2+a} \geq \dfrac{(\displaystyle \sum a)^2}{\displaystyle \sum a^2+\sum a},$ which rewrites as $2k=2\sum ab \leq \sum a$. Thus, to prove the desired inequality it suffices to prove that $3k+\dfrac{2}{k} \leq 5,$ that is $(k-1)(3k-2) \leq 0,$ or equivalently $\dfrac{2}{3} \leq k \leq 1,$ which by our Claim is true, and so we are done. Who is the author for this proof?
Orestis_Lignos
15.05.2023 14:44
Me, Orestis Lignos.
mihaig
16.05.2023 08:06
Thank you. Nice one