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Prove for all $a, b, c in mathbb {R^+}$ $${aoversqrt{a+b}} + {boversqrt{b+c}} + {coversqrt{c+a}} gt sqrt{a+b+c}$$ is true

I have been studying about inequalities and how to prove different inequalities, and I stumbled over this question which I failed to answer. I failed to see a connection between $sqrt{a+b+c} text{ and} {aoversqrt{a+b}} + {boversqrt{b+c}} + {coversqrt{c+a}}$. Is there any way I can prove this question and if there is can you please help. Thank you.

You may proceed as follows using convexity of $frac{1}{sqrt{x}}$:

• 
  $sum_{k=1}^3 lambda_k f(x_k) geq fleft( sum_{k=1}^3 lambda_k x_kright)$ with $sum_{k=1}^3 lambda_k = 1$, $lambda_1,lambda_2,lambda_3 geq 0$
  $$
  begin{eqnarray*} {aoversqrt{a+b}} + {boversqrt{b+c}} + {coversqrt{c+a}}
  & = & (a+b+c)sum_{cyc}frac{a}{(a+b+c)(a+b)} \
  & stackrel{Jensen}{geq} & (a+b+c)frac{1}{sqrt{frac{1}{a+b+c}sum_{cyc}a(a+b)}} \
  & color{blue}{>} & (a+b+c)frac{1}{sqrt{frac{1}{a+b+c}sum_{cyc}a(a+bcolor{blue}{+c})}} \
  & = & (a+b+c)frac{1}{sqrt{frac{1}{a+b+c}(a+b+c)^2}} \
  & = & color{blue}{sqrt{a+b+c}}
  end{eqnarray*}
  $$
  

You need only Jensen applied to convex function $f(x)={1 over sqrt{x}}$:

$$LHS=aspace f(a+b)+bspace f(b+c)+cspace f(c+a)ge(a+b+c)space f(frac{a(a+b)+b(b+c)+c(c+a)}{a+b+c})$$

$$LHSge(a+b+c)space f(frac{(a+b+c)^2-ab-bc-ac}{a+b+c})$$

$$LHSge(a+b+c)sqrt{frac{a+b+c}{(a+b+c)^2-ab-bc-ac}}$$

$$LHSgt(a+b+c)sqrt{frac{a+b+c}{(a+b+c)^2}}$$

$$LHSgtsqrt{a+b+c}$$

Holder's Inequality: Let $a_{ij}, 1 leq i leq m , 1 leq j leq m $ be positive real numbers. Then the following inequality holds:

$$prod_{i = 1}^{m}left(sum_{j = 1}^{n}a_{ij}right) geq
left(sum_{j = 1}^{n} sqrt[m]{prod_{i = 1}^{m} a_{ij}}right)^{m}.
> $$

For example, consider real numbers $a, b, c, p, q, r, x, y, z$. Then,

$$(a^{3} + b^{3} + c^{3})(p^{3} + q^{3} + r^{3})(x^{3} + y^{3} +
> z^{3}) geq (aqx + bqy + crz)^{3}. $$

We have

$$sum_{text{cyc}} frac{a}{sqrt{a + b}} > sum_{text{cyc}} frac{a}{sqrt{a + 2b}}. $$

By Holder's Inequality,

$$left(sum_{text{cyc}} frac{a}{sqrt{a + 2b}}right)left(sum_{text{cyc}} frac{a}{sqrt{a + 2b}}right)left(sum_{text{cyc}} a(a + 2b)right) geq (a + b+ c)^{3}. $$

Thus,

$$left(sum_{text{cyc}}frac{a}{sqrt{a + 2b}}right)^{2} geq a + b + c, $$

and the result follows.

too long; can't fit into a comment. I think @Ekesh 's answer has a little flaw: @Ekesh please see if mine is correct.

Your inequality is equivalent to

$$sum_{text{cyc}} frac{a}{sqrt{a + b}}geq a+b+c$$

By Holder's,

$$left(sum_{text{cyc}} frac{a}{sqrt{a + b}}right)left(sum_{text{cyc}} frac{a}{sqrt{a + b}}right)left(sum_{text{cyc}} a(a + b)right) geq (a + b+ c)^{3}.$$

We notice that (by expanding),

$$left(sum_{text{cyc}} a(a + b)right) leq (a + b+ c)^{2}.$$

It is valid to divide the above equation with this as $fbox{1}$ It is positive $fbox{2}$ This move removes more value at the RHS, which will not affect the sign.

Hence we get

$$left(sum_{text{cyc}}frac{a}{sqrt{a + b}}right)^{2} geq a + b + c,$$

which is what we want.