Let $a,b,c,d$ be positive real numbers such that $a^2+b^2+c^2+d^2=4$. Prove that exist two of $a,b,c,d$ with sum less or equal to $2$.
Problem
Source: Greece JBMO TST 2018 p1
Tags: algebra, inequalities, four variables
sqing
29.04.2019 03:05
parmenides51 wrote: Let $a,b,c,d$ be positive real numbers such that $a^2+b^2+c^2+d^2=4$. Prove that exist two of $a,b,c,d$ with sum less or equal to $2$. https://artofproblemsolving.com/community/c6h1621128p10145829
fabijan_123
31.01.2020 17:01
We get solution by contradiction. So, by contradiction se say that: a + b > 2, a + c > 2, ... Because of that, a + b + c + d > 4. By AM-QM we get contradiction because LHR > RHR. So, there exist two potisive numbers which sum is less or equal 2.
Keith50
17.08.2020 11:43
WLOG, let $a\geq b\geq c\geq d$, assume for contradiction that $d\leq 1, a,b,c \geq 1$ and any sum of two numbers are greater than $2$.(if $a,b,c,d\geq 1$, then it is obvious that equality holds at $a=b=c=d=1$) By Cauchy, we have \[16=(a^2+b^2+c^2+d^2)(c^2+d^2+a^2+b^2)\geq (2ac+2bd)^2\implies 2\geq ac+bd\geq c+d\]which contradicts our assumption, so there must exist a sum of two numbers that are less than or equal to $2.\ \square$
hiker
09.09.2020 20:52
By Cauchy-Swarz, we have
$$2(a^2+b^2) \ge (a+b)^2$$$$2(c^2+d^2) \ge (c+d)^2$$Adding the two inequalities, we have
$$2(a^2+b^2+c^2+d^2)=2\cdot4=8 \ge (a+b)^2 + (c+d)^2$$
If $a+b, c+d > 2$, then $(a+b)^2 + (c+d)^2 > 2^2 + 2^2 = 8$, contradiction. Hence, one of $a+b, c+d$ is less or equal to $2$.
vulam
09.09.2020 20:57
parmenides51 wrote: Let $a,b,c,d$ be positive real numbers such that $a^2+b^2+c^2+d^2=4$. Prove that exist two of $a,b,c,d$ with sum less or equal to $2$. umm is the answer 1? like a=1 b=1 c=1 d=1
vulam
09.09.2020 20:58
im wrong
OlympusHero
11.10.2021 00:53
I claim that $(a+b)^2+(a+c)^2+(a+d)^2+(b+c)^2+(b+d)^2+(c+d)^2 \leq 24$. Expanding gives that it suffices to show $2(ab+ac+ad+bc+bd+cd) \leq 12$, or $ab+ac+ad+bc+bd+cd \leq 6$. But we have $(a-b)^2+(a-c)^2+(a-d)^2+(b-c)^2+(b-d)^2+(c-d)^2 \geq 0$, or $3(a^2+b^2+c^2+d^2) \geq 2(ab+ac+ad+bc+bd+cd) \implies ab+ac+ad+bc+bd+cd \leq 6$, as desired. Now assume all pairs from $a,b,c,d$ have sum greater than $2$. But this means $(a+b)^2, (a+c)^2, (a+d)^2, (b+c)^2, (b+d)^2, (c+d)^2 > 4$, and adding these gives a contradiction. All steps are reversible, so we're done.
Mahdi_Mashayekhi
08.01.2022 17:09
(a^2 + b^2 + c^2 + d^2)(1 + 1 + 1 + 1) ≥ (a + b + c + d)^2 ---> 16 ≥ (a + b + c + d)^2 ---> 4 ≥ (a + b + c + d) so one of a+b or c+d is equal or less than 2.
huashiliao2020
28.03.2023 07:06
Suppose for the sake of contradiction that each of the 4 choose 2 pairs’ sums are all >2. Then each variable appears 3 times in a pair (take any var. and then you have three options left), and summing up all equations we have it is >6x2=12. However, each variable appears three times, so we have a+b+c+d>4. By AM-QM, $a+b+c+d\leq4\sqrt{(a^2+b^2+c^2+d^2)/4}=4\sqrt{4/4}=4$, contradiction.
ryanbear
29.07.2023 09:18
By cauchy, $(1+1+1+1)(a^2+b^2+c^2+d^2) = 16 \ge (a+b+c+d)^2$ $a+b+c+d \le 4$ Proof by contradiction, if any sum of two integers is greater than $2$, then $a+b > 2$ and $c+d > 2$, so $a+b+c+d>4$, contradiction So there exists two of $a,b,c,d$ with sum less than or equal to two
Jishnu4414l
01.05.2024 11:39
FTSOC let, $a+b>2$, $b+c>2$, $c+d>2$, and $d+a>2$. Now square and add these inequalities to get $2\sum_{cyc}{a^2}+2\sum_{cyc}{ab}>16$ $\implies \sum_{cyc}{ab}> 4$ However we have, $\sum_{cyc}{ab}\leq\sum_{cyc}{a^2}=4$. This is a contradiction. So atleast one of these $4$ inequalities must not hold, and we're done!