Question:
let $a,b,c>0$ show that $$(\sqrt{a}+\sqrt{b}+\sqrt{c})^2(a+b+c)^3\ge 27(ab+bc+ac)^2$$
since $$(a+b+c)^2\ge 3(ab+bc+ac)\Longleftrightarrow (a-b)^2+(b-c)^2+(c-a)^2\ge 0$$
it is enough to show that $$(\sqrt{a}+\sqrt{b}+\sqrt{c})^2(a+b+c)\ge 9(ab+bc+ac)$$ Following it hard to prove.maybe this inequality can use Cauchy-Schwarz inequality solve it?
Your last inequality is wrong. Try $c\rightarrow0^+$ and $a=b=1$.
But your starting inequality is true.
We need to prove that $$(a+b+c)^2(a^2+b^2+c^2)^3\geq27(a^2b^2+a^2c^2+b^2c^2)^2$$ for positives $a$, $b$ and $c$.
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, we need to prove that $$u^2(3u^2-2v^2)^3\geq(3v^4-2uw^3)^2$$ or $$u\sqrt{(3u^2-2v^2)^3}-3v^4+2uw^3\geq0.$$ Thus, it remains to prove our inequality for a minimal value of $w^3$.
Now, we know that $a$, $b$ and $c$ are positive roots of the equation $$(x-a)(x-b)(x-c)=0$$ or $$x^3-3ux^2+3v^2x-w^3=0$$ or $$w^3=x^3-3ux^2+3v^2x,$$ which says that a line $y=w^3$ and the graph of $y=x^3-3ux^2+3v^2x$ have three common points and $w^3$ gets a minimal value, when the line $y=w^3$ is a tangent line to the graph of $y=x^3-3ux^2+3v^2x$, which happens for equality case of two variables.
Also we must check the case $w^3\rightarrow0^+$.
Thus, it's enough to prove our inequality in the following cases.
Let $c\rightarrow0^+$ and $b=1$.
We have here $(a+1)^2(a^2+1)^3\geq27a^4$, which is AM-GM: $$(a+1)^2(a^2+1)^3\geq\left(2\sqrt{x}\right)^2(2a)^3=32a^4\geq27a^4;$$
Also we can make the following.
We need to prove that $$(\sqrt{a}+\sqrt{b}+\sqrt{c})^2(a^2+b^2+c^2)(a+b+c)^3\geq27(ab+ac+bc)^2(a^2+b^2+c^2).$$ Now, by Holder $$(\sqrt{a}+\sqrt{b}+\sqrt{c})^2(a^2+b^2+c^2)\geq(a+b+c)^3.$$ Thus, it remains to prove that $$(a+b+c)^6\geq27(ab+ac+bc)^2(a^2+b^2+c^2),$$ which is AM-GM: $$(a+b+c)^6=(a^2+b^2+c^2+(ab+ac+bc)+(ab+ac+bc))^3\geq$$ $$\geq\left(3\sqrt[3]{(a^2+b^2+c^2)(ab+ac+bc)^2}\right)^3=27(ab+ac+bc)^2(a^2+b^2+c^2).$$ Done!