Articles of rearrangement inequality

How to prove that $\frac x{\sqrt y}+\frac y{\sqrt x}\ge\sqrt x+\sqrt y$

I am trying to prove that $$\frac x{\sqrt y}+\frac y{\sqrt x}\ge\sqrt x+\sqrt y$$ I tried some manipulations, like multiplications in $\sqrt x$ or $\sqrt y$ or using $x=\sqrt x\sqrt x$, but I’m still stuck with that. What am I missing?

Inequality: $ab^2+bc^2+ca^2 \le 4$, when $a+b+c=3$.

Let $a,b,c $ are non-negative real numbers, and $a+b+c=3$. How to prove inequality $$ ab^2+bc^2+ca^2\le 4.\tag{*} $$ In other words, if $a,b,c$ are non-negative real numbers, then how to prove inequality $$ 27(ab^2+bc^2+ca^2)\le 4(a+b+c)^3.\tag {**} $$ $\color{gray}{\mbox{(Without using “universal” Lagrange multipliers method).}}$ Thanks!

Algebraic proof that if $a>0$ then $1+a^9 \leq \frac{1}{a}+a^{10} $

Prove that if $a>0$ then $1+a^9 \leq \frac{1}{a}+a^{10} $. Using the fact that $a \gt 0$, multiply by $a$ on both sides and get everything to one side we have; $a^{11}-a^{10}-a+1 \geq 0$. By factoring $(a^{10}-1)(a-1) \geq 0 $. I am not sure how to proceed any further.

Proving the inequality $4\ge a^2b+b^2c+c^2a+abc$

So, a,b,c are non-negative real numbers for which holds that $a+b+c=3$. Prove the following inequality: $$4\ge a^2b+b^2c+c^2a+abc$$ For now I have only tried to write the inequality as $$4\left(\frac{a+b+c}3\right)^3\ge a^2b+b^2c+c^2a+abc$$ but I don’t know what to do after that…

Proof of the inequality $\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b} \geq \frac{3}{2}$

I am currently attempting to prove the following inequality $\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b} \geq \dfrac{3}{2}$ for all $ a,b,c>0$ My instinctive plan of attack is to use the AM/GM inequality with $x_1=\dfrac{a}{b+c}$ etc. Using that I get this $\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b} \geq 3\times \sqrt[3]{\dfrac{abc}{(a+b)(b+c)(a+c)}}$ From here, I used the fact that $(a+b)(b+c)(a+c)\geq 8abc$, which can be easily proven by considering […]