Show that for any non-negative real numbers $a,b,c,d$ such that $a^2+b^2+c^2+d^2=1$ the following inequality hold: $$a+b+c+d-1\geq 16abcd$$When does equality hold?
Problem
Source: Kosovo MO 2019 Grade 12, Problem 3
Tags: inequalities, algebra, Kosovo
03.03.2019 01:41
dangerousliri wrote: Show that for any non-negative real numbers $a,b,c,d$ such that $a^2+b^2+c^2+d^2=1$ the following inequality hold: $$a+b+c+d-1\geq 16abcd$$When does equality hold? Let $t=\sqrt{abcd}$. Using AM-GM inequality, we get $ 0 \leq t \leq \frac{1}{4}$. The inequality is equivalent to prove $$(a+b+c+d)^2 \geq (1+16abcd)^2$$or equivalently $$ab+bc+cd+da+ac+bd\geq 128t^4+16t^2$$But according to AM-GM inequality $$ab+bc+cd+da+ac+bd\geq 6t$$Therefore, it is enough to prove $$3t \geq 64t^4+8t^2$$which we can rewrite as $$t (1-4 t) (16 t^2 + 4 t + 3) \geq 0$$which is obviously true. Equality holds for $(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2})$ or $(1,0,0,0)$ and their permutations.
03.03.2019 02:54
dangerousliri wrote: Show that for any non-negative real numbers $a,b,c,d$ such that $a^2+b^2+c^2+d^2=1$ the following inequality hold: $$a+b+c+d-1\geq 16abcd$$When does equality hold? I'm glad to inform you that I'm the author of the problem ,among the professor Valmir B. Krasniqi from the University from Pristina. Of course, this problem has been created for a competition. But the below one is for study.
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03.03.2019 03:01
bel.jad5 wrote: dangerousliri wrote: Show that for any non-negative real numbers $a,b,c,d$ such that $a^2+b^2+c^2+d^2=1$ the following inequality hold: $$a+b+c+d-1\geq 16abcd$$When does equality hold? which is obviously true. Equality holds for $(1,1,1,1)$ or $(1,0,0,0)$ and their permutations. And what about $a=b=c=d=0.5$
03.03.2019 03:03
fixed...I meant when $a=b=c=d$.
03.03.2019 03:30
bel.jad5 wrote: dangerousliri wrote: Show that for any non-negative real numbers $a,b,c,d$ such that $a^2+b^2+c^2+d^2=1$ the following inequality hold: $$a+b+c+d-1\geq 16abcd$$When does equality hold? Let $t=\sqrt{abcd}$. Using AM-GM inequality, we get $ 0 \leq t \leq \frac{1}{4}$. The inequality is equivalent to prove $$(a+b+c+d)^2 \geq (1+16abcd)^2$$or equivalently $$ab+bc+cd+da+ac+bd\geq 128t^4+16t^2$$But according to AM-GM inequality $$ab+bc+cd+da+ac+bd\geq 6t$$Therefore, it is enough to prove $$3t \geq 64t^4+8t^2$$which we can rewrite as $$t (1-4 t) (16 t^2 + 4 t + 3) \geq 0$$which is obviously true. Equality holds for $(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2})$ or $(1,0,0,0)$ and their permutations. Both of us thank you for your beautiful solution. Among its beauty,the solution certifies that the problem has been an excellent choice for that level of competition.
04.03.2019 14:59
dangerousliri wrote: Show that for any non-negative real numbers $a,b,c,d$ such that $a^2+b^2+c^2+d^2=1$ the following inequality hold: $$a+b+c+d-1\geq 16abcd$$When does equality hold? See here, the official site: http://kms-ks.org/wp-content/uploads/2019/03/Klasa-XII.pdf http://kms-ks.org/omk/?fbclid=IwAR3DDRa_f_v0jPUC6uekpvTIsYbmMEvrAVOSLGnJ7Kv531YXO811Ynvw9-o
04.03.2019 18:55
dangerousliri wrote: Show that for any non-negative real numbers $a,b,c,d$ such that $a^2+b^2+c^2+d^2=1$ the following inequality hold: $$a+b+c+d-1\geq 16abcd$$When does equality hold? By AM-GM, We have $$\sqrt{abcd} \le \frac{a^2+b^2+c^2+d^2}{4}=\frac{1}{4}$$Since $a,b,c,d \le 1$, then $$a+b+c+d \ge a^2+b^2+c^2+d^2=1$$, By AM-QM $$a+b+c+d \le 4\sqrt{\frac{a^2+b^2+c^2+d^2}{4}}=2$$So we can write $$(2-(a+b+c+d)(a+b+c+d-1) \ge 0$$$$\iff a+b+c+d \ge \frac{2+(a+b+c+d)^2}{3}=1+\frac{2(ab+bc+cd+da+ac+bd)}{3}$$But according to AM-GM inequality $$ab+bc+cd+da+ac+bd \ge 6\sqrt{abcd}{}$$Then $$ a+b+c+d \ge 1+4\sqrt{abcd} \ge 1+16abcd$$Done.
04.03.2019 19:06
AM-QM ?What is it?
04.03.2019 19:10
Arithmetic Quadratic mean inequaity. See here https://www.geogebra.org/m/Fxu9tWwJ
04.03.2019 19:21
ali3985 wrote: dangerousliri wrote: Show that for any non-negative real numbers $a,b,c,d$ such that $a^2+b^2+c^2+d^2=1$ the following inequality hold: $$a+b+c+d-1\geq 16abcd$$When does equality hold? By AM-GM, We have $$\sqrt{abcd} \le \frac{a^2+b^2+c^2+d^2}{4}=\frac{1}{4}$$Since $a,b,c,d \le 1$, then $$a+b+c+d \ge a^2+b^2+c^2+d^2=1$$, By AM-QM $$a+b+c+d \le 4\sqrt{\frac{a^2+b^2+c^2+d^2}{4}}=2$$So we can write $$(2-(a+b+c+d)(a+b+c+d-1) \ge 0$$$$\iff a+b+c+d \ge \frac{2+(a+b+c+d)^2}{3}=1+\frac{2(ab+bc+cd+da+ac+bd)}{3}$$But according to AM-GM inequality $ab+bc+cd+da+ac+bd \ge 4\sqrt{abcd}{}$, then $$\iff a+b+c+d \ge 1+4\sqrt{abcd} \ge 1+16abcd$$Done. Bravo!
24.03.2019 00:38
dangerousliri wrote: Show that for any non-negative real numbers $a,b,c,d$ such that $a^2+b^2+c^2+d^2=1$ the following inequality hold: $$a+b+c+d-1\geq 16abcd$$When does equality hold? We have beautiful solutions here and the official solution. Soon,the author's solution.
31.05.2019 11:14
Interesting and engaging AM-GM gives $$\frac{a^2+b^2+c^2+d^2}{4}\ge \sqrt{abcd}\implies 1\ge 4 \sqrt{abcd}$$By AM-QM $$\sqrt{\frac{a^2+b^2+c^2+d^2}{4}}\ge\frac{a+b+c+d}{4}\implies 2\ge a+b+c+d$$$$\frac{a+b+c+d-1}{2} =\frac12\cdot \frac{(a+b+c+d)^2-1}{(a+b+c+d)+1}= \frac12\cdot \frac{(a+b+c+d)^2-(a^2+b^2+c^2+d^2)}{(a+b+c+d)+1}=$$$$=\frac{ab+ac+ad+bc+bd+cd}{(a+b+c+d)+1}\ge \frac{ab+ac+ad+bc+bd+cd}{3}$$AM-GM and first inequality in this post yield $$\frac{ab+ac+ad+bc+bd+cd}{3}\ge 2\sqrt[6]{a^3b^3c^3d^3}=2\sqrt{abcd}\ge2\sqrt{abcd}\cdot4 \sqrt{abcd}=8abcd$$Done!
31.05.2019 11:42
Bravo!!!
31.05.2019 17:29
Show that for any non-negative real numbers $a,b,c$ such that $a^2+b^2+c^2=3$ the following inequality hold: $$a+b+c-1\geq 2abc$$here
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