RMO Inequalities - 3

Applying $RMS \geq AM$

$\large{ \sqrt{\dfrac{a^2+b^2+c^2+d^2}{4}} \geq \dfrac{a+b+c+d}{4} \\ \Rightarrow \dfrac{a^2+b^2+c^2+d^2}{4} \geq \left( \dfrac{1}{4} \right)^2 \\ \Rightarrow a^2+b^2+c^2+d^2 \geq \dfrac{1}{4}\times \dfrac{1}{4}\times 4 = \dfrac{1}{4} = \boxed{0.25}}$

Apply Cauchy Schwartz inequality on $<a,b,c,d>$ and $<1,1,1,1>$ we get: $(a+b+c+d)^2\leq(a^2+b^2+c^2+d^2)(1+1+1+1)\\1^2\leq(a^2+b^2+c^2+d^2)4\\\implies(a^2+b^2+c^2+d^2)\geq\dfrac14=\boxed{0.25}$

Note: You can generalize this to the $n^{th}$ term, i.e. $\displaystyle\sum^n_{i=1}a_i^2\geq\dfrac1n$

By Titu's Lemma, $\frac{{a}^{2}}{1} + \frac{{b}^{2}}{1} + \frac{{b}^{2}}{1} + \frac{{d}^{2}}{1} \ge \frac{{(a+b+c+d)}^{2}}{4}$

And then we are done.

By the AM-GM inequality, a^2+1/16 >=mod(a/2)>=a/2. Similarly, this can be done for the other quantities. Adding up, we get a^2+b^2+c^2+d^2+1/4>=(a+b+c+d)/2=1/2, implying a^2+b^2+c^2+d^2>=1/4. Hence the minimum is 1/4, with equality holding if a=b=c=d=1/4.