We can simply use Cauchy-Schwarz inequality: $$ \frac{x}{y+z}+\frac{25y}{z+x}+\frac{4z}{x+y}=\frac{x^2}{xy+xz}+\frac{25y^2}{yz+xy}+\frac{4z^2}{xz+yz}\ge \frac{(x+5y+2z)^2}{2(xy+yz+zx)}>2, $$because $(x+5y+2z)^2=(x^2+25y^2+4z^2)+(10xy+20yz+4xz)>4(xy+yz+zx)$.

$$\sum_{cyc}{a\left(\frac{x}{y+z}\right)}= (x+y+z)\left(\sum_{cyc}{\frac{a}{y+z}}\right) - ( a+b+c)$$$$\implies \frac{1}{2}\left(\sum_{cyc}{y+z}\right) \left(\sum_{cyc}{\frac{a}{y+z}}\right)-(a+b+c) >= \frac{1}{2}{\left(\sum_{cyc}{\sqrt{a}}\right)}^2 - (a+b+c) \ ( \ By \ Cauchy-Schwarz \ inequality )$$Put $ a = 1, b=25,c=4 \ $ we get $$\sum_{cyc}{a\left(\frac{x}{y+z}\right)} >= 2 $$Checking for equality condition for Cauchy-Schwarz inequality we get $x < 0$ ( Contradiction ) $$\therefore \sum_{cyc}{a\left(\frac{x}{y+z}\right)} > 2 $$HENCE PROVED.

Prove that for $\forall$ $x,y,z\in \mathbb{R}^+$ the following inequality is true: $$\frac{x}{y+z}+\frac{25y}{z+x}+\frac{4z}{x+y}>3$$See also here https://artofproblemsolving.com/community/c6h1643160p10973643

richrow12 wrote: We can simply use Cauchy-Schwarz inequality: $$ \frac{x}{y+z}+\frac{25y}{z+x}+\frac{4z}{x+y}=\frac{x^2}{xy+xz}+\frac{25y^2}{yz+xy}+\frac{4z^2}{xz+yz}\ge \frac{(x+5y+2z)^2}{2(xy+yz+zx)}>2, $$because $(x+5y+2z)^2=(x^2+25y^2+4z^2)+(10xy+20yz+4xz)>4(xy+yz+zx)$. With this approach we even see that the constant in the inequality can be replaced by $3$, since $\frac{(x+5y+2z)^2}{2(xy+yz+zx)} > 3$ is equivalent to $x^2 + 4xy - 2xz + 25 y^2 + 14yz + 4z^2 > 0$, true due to $(x-z)^2 + 4xy + 25y^2 + 14yz + 3z^2 > 0$.

Let $a, b, c \geq 0$ and $ ab+bc+ca>0 . $ Prove that $$\frac{a}{b+c}+\frac{4b}{c+a}+\frac{9c}{a+b} \geq 4$$

Pinko wrote: Prove that for $\forall$ $x,y,z\in \mathbb{R}^+$ the following inequality is true: $\frac{x}{y+z}+\frac{25y}{z+x}+\frac{4z}{x+y}>2$. Let $a, b, c >0$ . Prove that $$\frac{a}{b+c}+\frac{b}{c+a}+\frac{4c}{a+b} \ge 2$$