Let us note $\sqrt{a}=x,\sqrt{b}=y,\sqrt{c}=z$ Left member turn into : $(x^2+y^2+z^2)^2\geq (x^2+y^2+z^2)(xy+yz+zx)$, which is obviously true ($x^2+y^2+z^2\geq xy+yz+zx$) Right member turn into: $(x^4+y^4+z^4)+(x^2+y^2+z^2)(xy+yz+zx)\geq 4(x^2y^2+y^2z^2+z^2x^2)$ . Here we appy 2nd deegre Schur :$(x^4+y^4+z^4+x^2yz+y^2zx+z^2xy)\geq (x^3y+y^3x)+(x^3z+z^3x)+(y^3z+z^3y)$ $(x^4+y^4+z^4+x^2yz+y^2zx+z^2xy)+(x^3y+y^3x)+(x^3z+z^3x)+(y^3z+z^3y)\geq 2((x^3y+y^3x)+(x^3z+z^3x)+(y^3z+z^3y)\geq 4(\sqrt {x^4y^4}+\sqrt {y^4z^4}+\sqrt{x^4x^4}=4(x^2y^2+y^2z^2+z^2x^2)$. Done.

Lets Start prove right side ! Our inequality is homogenous . we can suppose $a+b+c=1$ . Now we need prove : $$3(a^2+b^2+c^2) \geq (\sqrt{ab}+\sqrt{bc}+\sqrt{ca}) + 2(a^2+b^2+c^2) - 2(ab+ac+bc) \to a^2+b^2+c^2 \geq (\sqrt{ab}+\sqrt{bc}+\sqrt{ca})- 2(ab+ac+bc) $$Then : $$(a+b+c)^2 \geq \sqrt{ab}+\sqrt{bc}+\sqrt{ca} \to 1 \geq \sqrt{ab}+\sqrt{bc}+\sqrt{ca} $$ By AM-GM inequality we can write : $$ \sqrt{ab} \leq \frac{a+b}{2} \to \sqrt{ab}+\sqrt{bc}+\sqrt{ca} \leq 1$$$\blacksquare$

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