Inequality proof (strange)

Question

Given $a^2 +b^2 +c^2 +d^2 =1$ where $a,b,c,d$ are positive real numbers, prove that $a+b+c+d-1 \geq 16abcd$

How can I prove the inequality ?

My attempts:

By Cauchy-Schwarz : $ (a+b+c+d)^{2} \leq (a^2 +b^2 +c^2 +d^2 )\cdot (1^2 +1^2 +1^2 + 1^2 ) $ Or $ 0\leq a+b+c+d -1 \leq 1 $ .

By AM-GM : $ a^2 + b^2 +c^2 +d^2 \geq 4 \sqrt[4]{(abcd)^2 } $ or $ 16abcd \leq 1 $ What can I do from here ?

@Cat master I solved your problem. If you want to see my solution, show please us your attempts. — Michael Rozenberg, Mar 03 '19 at 20:48
I tried AM-GM , Cauchy Swarchz inequality ,...etc , but cannot solve it , can you post the solution and to save time post it in short-terms . — Cat master, Mar 03 '19 at 20:53
@Catmaster If you don't show work, it seems like you're just trying to get others to do homework for you. By showing your work, others get the opportunity to potentially show you where you might be going wrong in your attempts or what insights you might need. — Michael, Mar 03 '19 at 21:01
Check this out for properly formatting math expressions. Learning this for 10 minutes will benefit you for years to come. — Lee David Chung Lin, Sep 01 '19 at 03:48

score 0 · Accepted Answer · answered Mar 03 '19 at 20:55

0

Let $a+b+c+d=4u$, $ab+ac+bc+ad+bd+cd=6v^2$ and $u^2=tv^2$.

Thus, since $u^2\geq v^2$ it's $$\sum_{sym}(a-b)^2\geq0,$$ we get $t\geq1$ and since by AM-GM $$ab+ac+bc+ad+bd+cd\geq6\sqrt{abcd},$$ it's enough to prove that $$(a+b+c+d)\sqrt{(a^2+b^2+c^2+d^2)^3}-(a^2+b^2+c^2+d^2)^2\geq16v^4$$ or $$2u\sqrt{(4u^2-3v^2)^3}-(4u^2-3v^2)^2\geq v^4$$ or $$t(4t-3)^3\geq(8t^2-12t+5)^2$$ or $$(t-1)(48t^2-68t+25)\geq0$$ and we are done!

answered Mar 03 '19 at 20:55

Michael Rozenberg

194,933

Can down-voter explain us why did you do it? – Michael Rozenberg Sep 09 '19 at 12:53

Inequality proof (strange)

1 Answers1