If $x,y,z\gt 0$ and $xyz=1$ Then minimum value of $\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}$ 
If $x,y,z\gt 0$ and $xyz=1$ Then find the minimum value of $\displaystyle \frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}$

$\bf{My\; Try::}$Using Titu's Lemma $$\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\ge \frac{(x+y+z)^2}{2(x+y+z)} = \frac{x+y+z}{2}\ge 3\frac{\sqrt[3]{xyz}}{2} = \frac{3}{2}$$
and equality holds when $$x=y=z=1$$
My question is how can we solve it without the above lemma, like using Jensen's Inequality or other inequality.
Please explain me.
Thanks

This answer shows how Nesbitt's inequality can used with other proof techniques to prove the given inequality.

Nesbitt's inequality
$$\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y} \geq \frac{3}{2}\tag{$*$}$$


$(1a)$ Observe that
$$\sum_{cyc}\left(\frac{x^2}{y+z}\right)+(x+y+z)=(x+y+z)\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\right)\tag{A}$$
Using $(*)$ and (A), we get
$$\sum_{cyc}\left(\frac{x^2}{y+z}\right) \geq \frac{(x+y+z)}{2}\tag{B}$$
Now, using AM-GM inequality, we get
$$\frac{(x+y+z)}{2} \geq  \frac{3\sqrt[3]{xyz}}{2}=\frac{3}{2}\tag{C}$$
Using (B) and (C), we get the desired result.
$(1b)$ Observe that
$$\sum_{cyc}\left(\frac{x^2}{y+z}\right)=(x+y+z)\left[\left(\sum_{cyc}\frac{x}{y+z}\right)-1\right]\tag{D}$$
Using $(*)$ and (C) in (D), we get
$$\sum_{cyc}\left(\frac{x^2}{y+z}\right) \geq 3 \cdot \left(\frac{3}{2}-1\right)=\frac{3}{2}$$

$(2)$
Without loss of generality, let $x \geq y \geq z$
Therefore, $$\frac{x}{y+z} \geq \frac{y}{x+z} \geq \frac{z}{x+y}$$
Now, using the Chebyshev inequality for these two increasing sequences, we get
$$3 \sum_{cyc}\left(\frac{x^2}{y+z}\right) \geq (x+y+z)\sum_{cyc}\left(\frac{x}{y+z}\right)\tag{E}$$
Using $(*)$ and (C) in (E), we get the required result.

$(3)$ Let $f(x)=x^2$
Now, we can write
$$\sum_{cyc}\left(\frac{x^2}{y+z}\right)=\sum_{cyc}\left(\frac{f(x)}{y+z}\right)$$
Note, that $f''(x)=2 > 0$, so the function is convex.
Now, using the Weighted-Jensen inequality, we get
$$\sum_{cyc}\left(\frac{f(x)}{y+z}\right) \geq 3f(M)\tag{F}$$
where using $(*)$, we get
$$3M=\sum_{cyc}\left(\frac{x}{y+z}\right) \geq \frac{3}{2} \Longleftrightarrow M \geq \frac{1}{2}\tag{G}$$
Using (G) in (F), we get the desired result.
Note: Notice that this method is independent of the constraint $xyz=1$

Finally, note that equality is indeed attained when $\boxed{x=y=z=1}$
