if a,b ∈ R >>> sqrt(a^2 + b^2)>= (|a|+|b|)/sqrt(2)

sqrt(a^2+(1-a^2))+sqrt(b^2+(1-b^2))+sqrt(c^2+(1-c^2))>=(3sqrt(2))/2??

please give some hints. I have no idea how to solve it.

if a,b>0 >>> (a^2+b^2)/(a+b)>=(a+b)/2
if a=b=0 >>> (a^2+b^2)/(a+b)=(a+b)/2

(a^4+b^4)/(a^2+b^2)+(b^4+c^4)/(b^2+c^2)+(c^4+a^4)/(c^2+a^2)>=ab+bc+ca, for any a,b,c>0

is it correct if I say:
(a^4+b^4)/(a^2+b^2)>=(a^2+b^2)/2
(b^4+c^4)/(b^2+c^2)>=(b^2+c^2)/2
(c^4+a^4)/(c^2+a^2)>=(c^2+a^2)/2
____________________________
(a^4+b^4)/(a^2+b^2)+(b^4+c^4)/(b^2+c^2)+(c^4+a^4)/(c^2+a^2)>=[2(a^2+b^2+c^2)]/2
(a^4+b^4)/(a^2+b^2)+(b^4+c^4)/(b^2+c^2)+(c^4+a^4)/(c^2+a^2)>=(a^2+b^2+c^2)
(a^4+b^4)/(a^2+b^2)+(b^4+c^4)/(b^2+c^2)+(c^4+a^4)/(c^2+a^2)>=[(a+b+c)^2]-2(ab+bc+ca)
(a^4+b^4)/(a^2+b^2)+(b^4+c^4)/(b^2+c^2)+(c^4+a^4)/(c^2+a^2)>=(a+b+c)^2>=2(ab+bc+ca)

??

I think I got it:
sqrt(a^2+(1-a^2))>=(|a|+|1-a|)/sqrt(2)
sqrt(b^2+(1-b^2))>=(|b|+|1-b|)/sqrt(2)
sqrt(c^2+(1-c^2))>=C

(|a|+|1-a|)/sqrt(2)>=|a+1-a|/sqrt(2)=2/sqrt(2)
>>>(|b|+|1-b|)/sqrt(2)>=2/sqrt(2) & (|b|+|1-b|)/sqrt(2)>=2/sqrt(2)
>>>sqrt(a^2+(1-a^2))+sqrt(b^2+(1-b^2))+sqrt(c^2+(1-c^2))>=(3sqrt(2))/2

let me know what you think

It would really help if you learned to use the LaTeX facility. And I have no idea of what you are trying to do.


is it correct if i say:
\frac{(a^4+b^4)}{(a^2+b^2)}\geq \frac{(a^2+b^2)}{2} \\
\frac{(b^4+c^4)}{(b^2+c^2)}\geq \frac{(b^2+c^2)}{2} \\
\frac{(c^4+a^4)}{(c^2+a^2)} \geq \frac{(c^2+a^2)}{2}
____________________________
\frac{(a^4+b^4)}{(a^2+b^2)}+\frac{(b^4+c^4)}{(b^2+ c^2)}+\frac{(c^4+a^4)}{(c^2+a^2)}\geq \frac{2(a^2+b^2+c^2)}{2} \\
\frac{(a^4+b^4)}{(a^2+b^2)}+\frac{(b^4+c^4)}{(b^2+ c^2)}+\frac{(c^4+a^4)}{(c^2+a^2)} \geq (a^2+b^2+c^2) \\
\frac{(a^4+b^4)}{(a^2+b^2)}+\frac{(b^4+c^4)}{(b^2+ c^2)}+\frac{(c^4+a^4)}{(c^2+a^2)} \geq (a+b+c)^2-2(ab+bc+ca) \\
\frac{(a^4+b^4)}{(a^2+b^2)}+\frac{(b^4+c^4)}{(b^2+ c^2)}+\frac{(c^4+a^4)}{(c^2+a^2)} \geq (a+b+c)^2\geq 2(ab+bc+ca)

??

The last line is not good. You cannot introduce this inequality sign like that.

EDIT: See ebaines post. I didn't know that... well, it is plausible (the expression is still true/valid), but I didn't know we could just insert it like that... =/

Actually I think the last line is good - if you known that m-n>0 (which you know from from the line above it), then m>n.

I don't understand this one, from post #2:
if a=b=0 >>> (a^2+b^2)/(a+b)=(a+b)/2

If a= b = 0, then (a^2 + b^2)/(a+b) is undefined.