Let a,b,c, and d be real numbers such that a≥b≥c≥d and a+b+c+d=13a2+b2+c2+d2=43. Show that ab≥3+cd.
Problem
Source: PMO 2021
Tags: algebra, number theory, inequalities, Philippines, PMO
20.03.2021 13:06
2021 Philippine MO
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20.03.2021 17:14
any ideas?
20.03.2021 20:49
From the hypothesis 43−a2≥(13−a)23we get 5/2≤a≤4 and so 52≤d≤c≤b≤a≤4.We have (a+b)2+(c+d)2=43+2(ab+cd)⇔(13−c−d)2+(c+d)2=43+2(ab+cd)So we must prove (13−c−d)2+(c+d)2≥43+2(2cd+3).This is equivalent to (c−d)2≥(c+d−6)(20−c−d).So it suffices to prove c+d≤6⇔a+b≥7.The inequality (b−c)(b−d)≥0 gives 3b2+(c+d)2+2cd≥(b+c+d)2.Now by replacing c+d=13−a−b, cd=a2+ab+b2−13(a+b)+63we will get 3b2−2(13−a)b+a2−13a+63≥0.Solve this as a quadratic polynomial of b with the notation b≥13−a3 (a+3b≥a+b+c+d=13)You will obtain a≥b≥13−a+√(4−a)(2a−5)3.This leads to 7/2≤a≤4 and now the inequality a+13−a+√(4−a)(2a−5)3≥7is satisfied. Equality holds for (4,3,3,3).
21.03.2021 06:10
Doxuantrong wrote: From the hypothesis we can easily obtain (by AM - GM) 52≤d≤c≤b≤a≤4.We have (a+b)2+(c+d)2=43+2(ab+cd)⇔(13−c−d)2+(c+d)2=43+2(ab+cd)So we must prove (13−c−d)2+(c+d)2≥43+2(2cd+3).This is equivalent to (c−d)2≥(c+d−6)(20−c−d).So it suffices to prove c+d≤6⇔a+b≥7.The inequality (b−c)(b−d)≥0 gives 3b2+(c+d)2+2cd≥(b+c+d)2.Now by replacing c+d=13−a−b, cd=a2+ab+b2−13(a+b)+63we will get 3b2−2(13−a)b+a2−13a+63≥0.Solve this as a quadratic polynomial of b with the notation b≥13−a3 (a+3b≥a+b+c+d=13)You will obtain a≥b≥13−a+√(4−a)(2a−5)3.This leads to 7/2≤a≤4 and now the inequality a+13−a+√(4−a)(2a−5)3≥7is satisfied. why 52≤d≤c≤b≤a≤4. ?
21.03.2021 08:43
Let a≥b≥c≥d be real numbers such that a+b+c+d=13, a2+b2+c2+d2=43.Prove that for all k∈R minWhen does equality hold ?
21.03.2021 12:55
Nice inequality and solution.
21.03.2021 15:51
go_placidly wrote: Let a, b, c, and d be real numbers such that a \geq b \geq c \geq d and a+b+c+d = 13a^2+b^2+c^2+d^2=43. Show that ab \geq 3 + cd. There is a simplier proof. The key is, \begin{align*} a^2+b^2+c^2+d^2+6\left(ab+cd \right) & = \left(a+b+c+d\right)^2+2\left(a-c\right)\left(b-d\right)+2\left(a-d\right)\left(b-c\right) \\ & \ge \left(a+b+c+d\right)^2,\end{align*}which gives us ab+cd \ge 21, or (a+b)^2+(c+d)^2 \ge a^2+b^2+c^2+d^2+42, (a+b)^2+(13-a-b)^2 \ge 85, (a+b-6)(a+b-7) \ge 0, a+b \ge 7.Hence, ab-cd \ge ab - \frac{c^2+d^2}{2} =ab + \frac{a^2+b^2-43}{2} =\frac{(a+b)^2-43}{2} \ge 3.
21.03.2021 16:10
Let \frac{39}{10}\ge a \geq b \geq c \geq d\geq 0, a+b+c+d = 13,a^2+b^2+c^2+d^2=43. Show that ab-cd \ge \frac 7{2}.
21.03.2021 16:24
sqing wrote:
Dear sir, I really respect you. But if you think a problem/solution is beautiful then kindly upvote the problem/solution rather then writing this.
30.03.2021 16:01
IMO Shortlist 2005 problem A3: Four real numbers p, q, r, s satisfy p+q+r+s = 9 and p^{2}+q^{2}+r^{2}+s^{2}= 21. Prove that there exists a permutation \left(a,b,c,d\right) of \left(p,q,r,s\right) such that ab-cd \geq 2.
22.01.2022 11:13
First we show it suffices to prove that a+b\geq7. 2ab+43=(a+b)^2+c^2+d^2\geq 49+c^2+d^2\geq 49+2cd.Therefore, ab\geq 3+cd. We now prove that a+b\geq 7. Let a+b=7-t and a^2+b^2=s. We show that t\leq0. (Clearly t\leq1/2) We have c+d=6+t and c^2+d^2=43-s. We solve for a,b,c,d in terms of s,t and apply the condition that b\geq c to get (7-t)-\sqrt{2s-(7-t)^2}\geq (6+t)+\sqrt{86-2s-(6+t)^2}rearrange and square both sides (can also see here again that t\leq 1/2) (1-2t)^2\geq 1+2t-2t^2+2\sqrt{(2s-(7-t)^2)(86-2s-(6+t)^2)},3t(t-1)\geq \sqrt{(2s-(7-t)^2)(86-2s-(6+t)^2)}\geq0.Therefore, as t\leq1/2 and t(t-1)\geq0, we get that t\leq 0.
29.05.2023 05:09
KaiRain wrote: go_placidly wrote: Let a, b, c, and d be real numbers such that a \geq b \geq c \geq d and a+b+c+d = 13a^2+b^2+c^2+d^2=43. Show that ab \geq 3 + cd. There is a simplier proof. The key is, \begin{align*} a^2+b^2+c^2+d^2+6\left(ab+cd \right) & = \left(a+b+c+d\right)^2+2\left(a-c\right)\left(b-d\right)+2\left(a-d\right)\left(b-c\right) \\ & \ge \left(a+b+c+d\right)^2,\end{align*}which gives us ab+cd \ge 21, or (a+b)^2+(c+d)^2 \ge a^2+b^2+c^2+d^2+42, (a+b)^2+(13-a-b)^2 \ge 85, (a+b-6)(a+b-7) \ge 0, a+b \ge 7.Hence, ab-cd \ge ab - \frac{c^2+d^2}{2} =ab + \frac{a^2+b^2-43}{2} =\frac{(a+b)^2-43}{2} \ge 3. How do you think of \begin{align*} a^2+b^2+c^2+d^2+6\left(ab+cd \right) & = \left(a+b+c+d\right)^2+2\left(a-c\right)\left(b-d\right)+2\left(a-d\right)\left(b-c\right) \\ & \ge \left(a+b+c+d\right)^2,\end{align*}at first?
29.05.2023 06:43
qinghong wrote: How do you think of \begin{align*} a^2+b^2+c^2+d^2+6\left(ab+cd \right) & = \left(a+b+c+d\right)^2+2\left(a-c\right)\left(b-d\right)+2\left(a-d\right)\left(b-c\right) \\ & \ge \left(a+b+c+d\right)^2,\end{align*}at first? The motivation is finding a relation betwen a+b+c+d, a^2+b^2+c^2+d^2, ab and cd. We know: \begin{align*} \frac{(a+b+c+d)^2-(a^2+b^2+c^2+d^2)}{2} = \left(ab+cd \right) + \left(ac+bd \right) + \left(ad +bc \right) ,\end{align*}which suggests us to prove: \begin{align*} 3 \left(ab+cd \right) \ge \left(ab+cd \right) + \left(ac+bd \right) + \left(ad +bc \right). \end{align*}