3412(a2+b2) ≥ (a+b)2
These days there seem to be a lot of inequalities being discussed here...
This one is much simpler...
If a, b, c are positive,
Prove that :
\frac{b^2+c^2}{b+c}+\frac{a^2+c^2}{a+c}+\frac{a^2+b^2}{a+b}\geq a+b+c
I hope think one has not been done before!
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14 Answers
1this can be easily done by considering Titu's lemma
\sum_{i = 1}^{n}\frac{x_{i}^{2}}{a_{i}}\ge\frac{(\sum_{i = 1}^{n}x_{i})^{2}}{\sum_{i = 1}^{n}a_{i}}
if we seperate the terms
it will be like
a^2/a+c
+b^2/b+c
+c^2/a+c
+a^2/a+b
+b^2/a+b
+c^2/b+c
now according to lemma
this is ≥ (2a+2b+2c)^2/4a+4b+4c
hence prooved....
62Dont titu's and rearrangement just kill so many inequalities ;)
39there is one more called muirhead if u would have ever read, an awesome one seriously...
62wow the statement of muirhead on wikipedia seems to be a generalization of rearrangement....
But i guess it will take me more reading to understand it clearly
19just a simple solution from my side:
we know that AM of 2nd powers > 2nd power of AMs
=> (b2+c2)/b+c > {(b+c)/2}2
=> (b2+c2)/b+c > (b+c)/2
similarly, (c2 + a2)/c+a >(c+a)/2 and (a2 + b2)/c+a >(b+a)/2
hence we have (b2+c2)/b+c + (c2 + a2)/c+a + (a2 + b2)/c+a > a+b+c
................work done !!!
62well doubly stochastic matrix is one whose sum or each row and column is 1