Another symmetric inequality

My solution:Call the expression is $A$

Because of $a+b+c=1$ and $a;b;c\geq 0$ so we have $0\leq a;b;c \leq 1$ .

From that we can see that $a\geq a^{2};b\geq b^{2};c\geq c^{2}$

So that we have $\sum \sqrt{5a+4}=\sum \sqrt{a+4a+4}\geq \sum \sqrt{a^2+4a+4}=\sum a+2$

And $A\geq a+b+c+6= 7$

The equality holds when $a=b=0;c=1$ or it's permutation.

Let $a=\sqrt{5x+4}$ , $b=\sqrt{5y+4}$ , and $c=\sqrt{5z+4}$

Squaring each expression gives:

$a^2=5x+4$ , $b^2=5y+4$ and $c^2=5z+4$

$a^2+b^2+c^2$ is therefore $5(x+y+z)+12$ = $5(1)+12$ = $17$ Knowing that :

$(a+b+c)^2\le3(a^2+b^2+c^2)$

Plugging in the values gives:

$(a+b+c)\le3(17)$

the minimum is therefore $\sqrt{51}$

By observation, we can see the expession reach its minimum when $(x,y,z)=(1;0;0)$ and its permutations, so put in the values, we get $f(x,y,z)\geq 7$ , now we need to prove this statement is true

Without losing generality, we assume that $x\geq y\geq z$ , then we have $z \in [0;\frac{1}{3}]$ . It's easy to prove that $f(x;y;z)\geq f(x;z;z)$ , so now we consider $f(x;z;z)=\sqrt{5x+4}+2\sqrt{5z+4}$ The condition becomes $x+2z=1 \Rightarrow x=1-2z$ $\therefore f(x;z;z)=\sqrt{9-10z}+2\sqrt{5z+4}$ Now we need to show that $f(x;z;z)\geq 7$ $\Leftrightarrow \sqrt{9-10z}+2\sqrt{5z+4}\geq 7$ $\Leftrightarrow 225z^2-140z\leq 0 \ (True \ for \ every \ z\in[0;\frac{1}{3}])$ The $LHS$ decrease in the interval $[0;\frac{1}{3}]$ , and $f(0)>f\big(\frac{1}{3}\big)$ so the equality holds when $z=0 \Rightarrow y=0;x=1$ and its permutations

Apply Cauchy Schwarz Inequality .a1=1, b1=(5x+4)^(1÷2) and similarly for a2, b2,a3, b3 we get , (a1b1+a2b2+a3b3)^2 = < (a1^2 +a2^2 +a3^2)(b1^2 +b2^2 +b3^2) by substituting we get the desired result .note: in a1 "1" is the subscript.