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 ?
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!