$a, b,c $ are positive real numbers such that $a+b+c = 3$, prove that :$a\sqrt{a+3}+b\sqrt{b+3}+c\sqrt{c+3} \geq 6$ Any ideas ?

consider positive numbers $a_1,a_2,a_3,\ldots,a_n$ and $b_1,b_2,\ldots,b_n$. does the following in-equality holds and if it does then how to prove it $\left[(a_1+b_1)(a_2+b_2)\cdots(a_n+b_n)\right]^{1/n}\ge \left(a_1a_2\cdots a_n\right)^{1/n}+\left(b_1b_2\cdots b_n\right)^{1/n}$

Let $a_1,a_2,\dots,a_n$ $\in$ $\mathbb{R}^+$ and $a_1\cdot a_2\cdots a_n=1$, , prove that $(1+a_1) \cdot (1+a_2) \cdot \dots \cdot (1+a_n) \geq 2^n$ I have tried factorising but it just lead me to extremily complicated equation that were extremily difficult to understand… Could someone help me?

If $a,b,c$ are positive reals such that $abc=1$, then prove that $$\sqrt\frac{a}{a+8} + \sqrt\frac{b}{b+8} +\sqrt\frac{c}{c+8} \geq 1$$ I tried substituting $x/y,y/z,z/x$, but it didn’t help(I got the reverse inequality). Need some stronger inequality. Thanks.

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