If $ab+bc+ca+abc=4$, then $\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\leq 3\leq a+b+c$

Let $a,b,c$ be positive reals such that $ab+bc+ca+abc=4$.

Then prove

$\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\leq 3\leq a+b+c $

So high guys im a high schooler trying to solve this inequality. I did a few things such as try to set a> b>c and then try to generalize it but i couldn't make much progress at all. I would appreciate a clear solution to help me understand the problem in depth.

I did manage to generalize that

$\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\leq a+b+c $

By using the AM-GM inequality on each of a b and c however havent made much progress in proving that the LHS is less than 3 and the RHS is greater than 3.

AM-GM on $ab$, $bc$, $ca$ and $abc$ gives: $$1=\frac{ab+bc+ca+abc}{4}\ge\sqrt[4]{(abc)^3}\implies 1\ge abc$$ So, still using the first equation: $$ab+bc+ca\ge 3$$ Now, by AM-GM $$a^2+b^2\ge 2ab$$ $$b^2+c^2\ge 2bc$$ $$c^2+a^2\ge 2ac$$ Adding them all up and dividing by $2$ gives: $$a^2+b^2+c^2\ge ab+bc+ca$$ So now: $$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)\ge 3(ab+bc+ca)\ge 9$$ From which it follows that: $$a+b+c\ge 3$$ I'll work on the other inequality when I have time

Both inequalities we can prove by the Contradiction method.

Indeed, for the right inequality let $a+b+c<3$, $a=kx$, $b=ky$ and $c=kz$,

where $k>0$ and $x+y+z=3$.

Thus, $k(x+y+z)<3$, which gives $k<1$.

Hence, $4=ab+ac+bc+abc=k^2(xy+xz+yz)+k^3xyz<xy+xz+yz+xyz$,

which is contradiction because we'll prove now that $$4\geq xy+xz+yz+xyz$$ or $$\frac{4(x+y+z)^3}{27}\geq\frac{(x+y+z)(xy+xz+yz)}{3}+xyz$$ or $$\sum_{cyc}(4x^3+3x^2y+3x^2z-10xyz)\geq0,$$ which is true by Muirhead.

Also, we can use AM-GM. $$\sum_{cyc}x^3=x^3+y^3+z^3\geq3\sqrt[3]{x^3y^3z^3}=3xyz;$$ $$\sum_{cyc}(x^2y+x^2z)=x^2y+x^2z+y^2x+y^2z+z^2x+z^2y\geq6\sqrt[6]{x^6y^6z^6}=6xyz.$$ Thus, $$\sum_{cyc}(4x^3+3x^2y+3x^2z-10xyz)=$$ $$=4(x^3+y^3+z^3)+3(x^2y+x^2z+y^2x+y^2z+z^2x+z^2y)-30xyz\geq$$ $$\geq4\cdot3xyz+3\cdot6xyz-30xyz=0.$$

The left inequality.

Let $\sqrt{ab}+\sqrt{ac}+\sqrt{bc}>3$, $a=kx$, $b=ky$ and $c=kz$,

where $k>0$ and $\sqrt{xy}+\sqrt{xz}+\sqrt{yz}=3$.

Thus, $k(\sqrt{xy}+\sqrt{xz}+\sqrt{yz})>3$, which gives $k>1$ and $$4=ab+ac+bc+abc=k^2(xy+xz+yz)+k^3xyz>xy+xz+yz+xyz,$$ which is contradiction because we'll prove now that $$4\leq xy+xz+yz+xyz.$$ Indeed, let $\sqrt{xy}=r$, $\sqrt{xz}=q$ and $\sqrt{yz}=p$.

Thus, $p+q+r=3$ and we need to prove that $$p^2+q^2+r^2+pqr\geq4$$ or $$\frac{(p+q+r)(p^2+q^2+r^2)}{3}+pqr\geq\frac{4(p+q+r)^3}{27}$$ or $$\sum_{cyc}(5p^3-3p^2q-3p^2r+pqr)\geq0$$ or $$\sum_{cyc}(p^3-p^2q-p^2r+pqr)+2\sum_{cyc}(2p^3-p^2q-p^2r)\geq0,$$ which is true by Schur and Muirhead.

The inequality $\sum\limits_{cyc}(2p^3-p^2q-p^2r)\geq0$ we can prove also without Muirhead: $$\sum_{cyc}(2p^3-p^2q-p^2r)=\sum_{cyc}(p^3-p^2q-pq^2+q^3)=\sum_{cyc}(p-q)^2(p+q)\geq0.$$

Done!

Also, we can use the following trigonometric substitution.

Let $\sqrt{ab}=2\cos\gamma$, $\sqrt{ac}=2\cos\beta$ and $\sqrt{bc}=\cos\alpha$.

Thus, $$\cos^2\alpha+\cos^2\beta+\cos^2\gamma+2\cos\alpha\cos\beta\cos\gamma=1,$$ where $\alpha$, $\beta$ and $\gamma$ are measures of acute angles.

Hence, easy to see that $\alpha+\beta+\gamma=180^{\circ}$ and we need to prove that $$\cos\alpha+\cos\beta+\cos\gamma\leq\frac{3}{2}$$ and $$\sum_{cyc}\frac{\cos\alpha\cos\beta}{\cos\gamma}\geq\frac{3}{2},$$ which is obvious.

Done again!