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.

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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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 $a=b=c=d=0.5$

fixed...I meant when $a=b=c=d$.

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.

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

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.

AM-QM ?What is it?

Arithmetic Quadratic mean inequaity. See here https://www.geogebra.org/m/Fxu9tWwJ

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!

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.

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!

Bravo!!!

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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