How to prove $ a+b+c+\sqrt{bc}+\sqrt{ca}+\sqrt{ab}\ge 6$?

Question

Question. Prove $$ a+b+c+\sqrt{bc}+\sqrt{ca}+\sqrt{ab}\ge 6,$$ when $a,b,c\ge 0: ab+bc+ca+abc=4.$

My idea:

I've tried to use AM-GM as $$\bullet \sum \sqrt{ab}\ge 2\sum \frac{ab}{a+b}=2(ab+bc+ca)\sum \frac{1}{a+b}-2(a+b+c)=\frac{2(ab+bc+ca)}{a+b+c}\left(3+\sum \frac{c}{a+b}\right)-2(a+b+c)$$

$$\rightarrow \sum \sqrt{ab} \ge \frac{2(ab+bc+ca)}{a+b+c}\left[3+ \frac{(a+b+c^2}{2(ab+bc+ca)}\right]-2(a+b+c)=\frac{6(ab+bc+ca)}{a+b+c}-(a+b+c)$$ Hence, $$\bullet a+b+c+\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\ge \frac{6(ab+bc+ca)}{a+b+c}.$$ It's enough to prove $ab+bc+ca\ge a+b+c.$ Unfortunately, the last inequality is obviously wrong.

I hope someone can share better approachs. Thank you.

Answer 1

I am not sure if this is 100% correct so please don't take it as a full answer, Here is my approach: by AM-GM inequality:

$ 4 = ab + bc + ca + abc \leq \left(\frac{a+b}{2}\right)^{2} + \left(\frac{b+c}{2}\right)^{2} + \left(\frac{a+c}{2}\right)^{2} + \left(\frac{a+b+c}{3}\right)^{3} $

Then

$ 16\times 27 \leq 27\left[\left(a+b\right)^{2}+\left(b+c\right)^{2}+\left(a+c\right)^{2}\right] + 4\left(a+b+c\right)^{3} $

Some more algebra

$ 8\times 27 \leq 27\left[a^{2}+b^{2}+c^{2} + 2ab+2bc+2ac\right] + 2\left(a+c+b\right)^{3} $

and

$ 8\times 27=6^{3}\leq 27\left(a+b+c\right)^{2} + 2\left(a+b+c\right)^{3} $ So maybe you could compare the RHS with

$\left(a+b+c+\sqrt{bc}+\sqrt{ca}+\sqrt{ab}\right)^{3}$

and check if it is smaller.

Answer 2

Let $a=\frac{2x}{y+z}$ and $b=\frac{2y}{x+z},$ where $x$, $y$ and $z$ are non-negatives such that $xy+xz+yz\neq0$.

Thus, the condition gives $c=\frac{2z}{x+y}$ and by C-S,AM-GM and Schur we obtain: $$\sum_{cyc}(a+\sqrt{ab})=2\sum_{cyc}\left(\frac{a}{b+c}+\sqrt{\frac{ab}{(a+c)(b+c)}}\right)=$$ $$=2\sum_{cyc}\left(\tfrac{a}{b+c}+\tfrac{(a+b)\sqrt{ab(a+c)(b+c)}}{(a+b)(a+c)(b+c)}\right)\geq2\sum_{cyc}\left(\tfrac{a}{b+c}+\tfrac{(a+b)\sqrt{ab}(\sqrt{ab}+c)}{\prod\limits_{cyc}(a+b)}\right)=$$ $$=2\sum_{cyc}\left(\tfrac{a}{b+c}+\tfrac{(a+b)ab+(ac+bc)\sqrt{ab})}{\prod\limits_{cyc}(a+b)}\right)\geq2\sum_{cyc}\left(\tfrac{a}{b+c}+\tfrac{(a+b)ab+2abc)}{\prod\limits_{cyc}(a+b)}\right)=$$ $$=6+2\sum_{cyc}\left(\tfrac{a}{b+c}+\tfrac{(a+b)ab+2abc)}{\prod\limits_{cyc}(a+b)}-1\right)=$$ $$=6+\tfrac{2\sum\limits_{cyc}(a^3+a^2b+a^2c+abc+a^2b+a^2c+2abc-3a^2b-3a^2c-2abc)}{\prod\limits_{cyc}(a+b)}=$$ $$=6+2\frac{\sum\limits_{cyc}(a^3-a^2b-a^2c+abc)}{\prod\limits_{cyc}(a+b)}\geq6.$$

Answer 3

Another way.

Let $\sum\limits_{cyc}(a+\sqrt{ab})<6,$ $a=kx^2$, $b=ky^2$ and $c=kz^2$, where $k>0$ and $x$, $y$ and $z$ are non-negatives such that $$\sum_{cyc}(x^2+xy)=6.$$ Thus, $k<1$ and $$4=ab+ac+bc+abc=k^2\sum_{cyc}x^2y^2+k^3x^2y^2z^2<\sum_{cyc}x^2y^2+x^2y^2z^2,$$ which is a contradiction because we'll prove now that $$4\geq\sum_{cyc}x^2y^2+x^2y^2z^2.$$ Indeed, we need to prove that $$4\left(\frac{\sum\limits_{cyc}(x^2+xy)}{6}\right)^3+\sum_{cyc}x^2y^2\cdot\frac{\sum\limits_{cyc}(x^2+xy)}{6}+x^2y^2z^2$$ or $$\sum_{cyc}(x^6+3x^5y+3x^5z-3x^4y^2-3x^4z^2-2x^3y^3+6x^3y^2z+6y^3z^2x+9x^4yz-20x^2y^2z^2)\geq0,$$ which is true by Schur and Muirhead: $$\sum_{cyc}(x^6+3x^5y+3x^5z-3x^4y^2-3x^4z^2-2x^3y^3+6x^3y^2z+6y^3z^2x+9x^4yz-20x^2y^2z^2)\geq$$ $$\geq\sum_{cyc}(4x^5y+4x^5z-3x^4y^2-3x^4z^2-2x^3y^3+6x^3y^2z+6y^3z^2x+8x^4yz-20x^2y^2z^2)\geq0.$$

Answer 4

Substitute $x=\sqrt{bc},y=\sqrt{ac},z=\sqrt{ab}$ ,we get that $x^2+y^2+z^2+xyz=4$

and we will prove $x+y+z+\frac{xy}{z}+\frac{yz}{x}+\frac{xz}{y}\geqslant6$

First,we prove that $x+y+z+x^2+y^2+z^2\geqslant6$

$\begin{align} &4=x^2+y^2+z^2+xyz\\\\&\leqslant\frac{x^2+y^2}{2}(z+2)+z^2\\\\&\Rightarrow\frac{x^2+y^2}{2}\geqslant\frac{4-z^2}{z+2}=2-z \end{align}$

Similarly,$\begin{align*}\frac{x^2+z^2}{2}\geqslant2-y\end{align*} $ ;$\begin{align*}\frac{y^2+z^2}{2}\geqslant2-x\end{align*} $

Add three inequalities together,we know $x+y+z+x^2+y^2+z^2\geqslant6$

Then we prove $\frac{xy}{z}+\frac{yz}{x}+\frac{xz}{y}\geqslant x^2+y^2+z^2$

By $\text{Cauchy-Schwarz}$

$\begin{align*}1=\sum_{cyc}\frac{\frac{xy}{z}}{2+\frac{xy}{z}}=\sum_{cyc}\frac{(\frac{xy}{z})^2}{2\cdot\frac{xy}{z}+(\frac{xy}{z})^2}\geqslant\frac{(\underset{cyc}{\sum}\frac{xy}{z})^2}{2\underset{cyc}{\sum}\frac{xy}{z}+\underset{cyc}{\sum}(\frac{xy}{z})^2}=\frac{\underset{cyc}{\sum}(\frac{xy}{z})^2+2\underset{cyc}{\sum}x^2}{2\underset{cyc}{\sum}\frac{xy}{z}+\underset{cyc}{\sum}(\frac{xy}{z})^2}\end{align*}$

So $\frac{xy}{z}+\frac{yz}{x}+\frac{xz}{y}\geqslant x^2+y^2+z^2$

So $x+y+z+\frac{xy}{z}+\frac{yz}{x}+\frac{xz}{y}\geqslant x+y+z+x^2+y^2+z^2\geqslant6$

Answer 5

Property 1: $\;$If $\;a\geqslant0\,,\;b\geqslant0\;$ and $\;ab\leqslant4\;$ then $\;\big(a-2\big)^{\!2}\!+b\left(a^2+a-4\right)+b^2\left(1+a-a^2\right)\geqslant0\;.$

Proof:
If $\;0\leqslant a\leqslant\dfrac85\;,\;$ then it results that
$\big(a-2\big)^{\!2}\!+b\left(a^2+a-4\right)+b^2\left(1+a-a^2\right)=$

$=\big(a-2\big)^{\!2}\!+b\left(a^2+a-4\right)+\dfrac{b^2\left(a^2+a-4\right)^2}{4(a-2)^2}+b^2\left[1+a-a^2-\dfrac{\left(a^2+a-4\right)^2}{4(a-2)^2}\right]=$

$=\!\!\left[(a\!-\!2)+\dfrac{b\!\left(a^2\!+\!a\!-\!4\right)}{2(a\!-\!2)}\right]^2\!\!+\dfrac{b^2\!\left(-5a^4\!+\!18a^3\!-\!21a^2\!+\!8a\right)}{4(a\!-\!2)^2}\!=$

$=\left[(a\!-\!2)+\dfrac{b\left(a^2\!+\!a\!-\!4\right)}{2(a\!-\!2)}\right]^2\!+\dfrac{ab^2(a-1)^2(8-5a)}{4(a-2)^2}\geqslant0\;.$

If $\;a>\dfrac85\;,\;$ then $\;a^2+a-4>\dfrac{64}{25}+\dfrac85-4=\dfrac4{25}>0\;$ and

$\big(a-2\big)^{\!2}\!+b\left(a^2+a-4\right)+b^2\left(1+a-a^2\right)\underset{\overbrace{\text{ because }ab\leqslant4\;}}{\geqslant}$

$\geqslant\dfrac{a^2b^2(a-2)^2}{16}+\dfrac{ab^2}4\left(a^2+a-4\right)+b^2\left(1+a-a^2\right)=$

$=\dfrac{b^2}{16}\bigg[a^2(a-2)^2+4a\left(a^2+a-4\right)+16\left(1+a-a^2\right)\!\bigg]=$

$=\dfrac{b^2}{16}\bigg[a^4-8a^2+16\bigg]=\dfrac{b^2}{16}\left(a^2-4\right)^2\geqslant0\;.$

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$\;$
Property 2: $\,$If $\,a\!\geqslant\!0\,,\,b\!\geqslant\!0\,,\,c\!\geqslant\!0\,$ and $\,ab\!+\!bc\!+\!ac\!+\!abc\!=\!4$

then it results that $\;\begin{cases}\sqrt{ab}\geqslant\dfrac{ab(1+c)}2\\[3pt]\sqrt{bc}\geqslant\dfrac{bc(1+a)}2\\[3pt]\sqrt{ac}\geqslant\dfrac{ac(1+b)}2\end{cases}$

Proof:

It results that
$\sqrt{ab}\geqslant\sqrt{ab}-\dfrac{c\sqrt{ab}\left(\sqrt a-\sqrt b\right)^2}{2\left(\sqrt{ab}+2\right)}=$

$=\dfrac{2ab+4\sqrt{ab}-ac\sqrt{ab}-bc\sqrt{ab}+2abc}{2\left(\sqrt{ab}+2\right)}=$

$=\dfrac{\sqrt{ab}\big(4-ac-bc\big)+2ab\big(1+c\big)}{2\left(\sqrt{ab}+2\right)} \underset{\overbrace{\text{ because }ab+bc+ac+abc=4\;}}{=}$

$=\dfrac{\sqrt{ab}\big(ab+abc\big)+2ab\big(1+c\big)}{2\left(\sqrt{ab}+2\right)}=\dfrac{ab(1+c)\left(\sqrt{ab}+2\right)}{2\left(\sqrt{ab}+2\right)}=$

$=\dfrac{ab(1+c)}2\;.$

Analogously we can prove the other two inequalities.

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$\;$
By using the previous properties we will prove that $$a+b+c+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\geqslant6$$ when $\;a,b,c\geqslant0\,:\,ab+bc+ac+abc=4\;.$

Proof:

It results that $\;a+b+ab>0\;,\;$ otherwise $\;a+b+ab=0\;$ would imply that $\;a=b=0\;$ and $\;ab+bc+ac+abc=0\;$ which would be a contradiction.

From $\;ab+bc+ac+abc=4\;,\;$ we get that $\;c=\dfrac{4-ab}{a+b+ab}$
and ,$\;$ since $\;c\geqslant0\;,\;$ it follows that $\;ab\leqslant4\;.$

Moreover, it results that

$a+b+c+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}=$

$=a+b+\dfrac{4-ab}{a+b+ab}+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}=$

$=\dfrac{a^2+ab+a^2b+b^2+ab^2+4}{a+b+ab}+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}=$

$=\dfrac{\big(a\!-\!2\big)^{\!2}\!+b\left(a^2\!+\!a\!-\!4\right)+b^2\!\left(1\!+\!a\!-\!a^2\right)+4a+4b+a^2b^2}{a+b+ab}+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\underset{\overbrace{\text{ by using Property 1 }}}{\geqslant}$

$\geqslant\dfrac{4a+4b+a^2b^2}{a+b+ab}+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\underset{\overbrace{\text{ by using Property 2 }}}{\geqslant}$

$\geqslant\dfrac{4a+4b+a^2b^2}{a+b+ab}+\dfrac{ab(1+c)}2+\dfrac{bc(1+a)}2+\dfrac{ac(1+b)}2=$

$=\dfrac{4a+4b+a^2b^2}{a+b+ab}+\dfrac{ab+bc+ac+abc+2abc}2=$

$=\dfrac{4a+4b+a^2b^2}{a+b+ab}+\dfrac{4+2abc}2=$

$=\dfrac{4a+4b+a^2b^2}{a+b+ab}+abc+2=$

$=\dfrac{4a+4b+a^2b^2}{a+b+ab}+\dfrac{ab(4-ab)}{a+b+ab}+2=$

$=\dfrac{4a+4b+a^2b^2+4ab-a^2b^2}{a+b+ab}+2=$

$=\dfrac{4(a+b+ab)}{a+b+ab}+2=4+2=6\;.$

Answer 6

Some thoughts with symmetric polynomials and calculus...

Let $a=x^2$, $b=y^2$, $c=z^2$ and $e_1=x+y=z$, $e_2=xy+yz+zx$, $e_3=xyz$.

The problem is: Minimize $e_1^2-e_2$ subject to $e_2^2-2e_1e_3+e_3^2=4$. The method of Lagrange's multpliers gives no solution. So the minumum occurs at the boundary $J=0.$

Jacobian $J$,$$J(e_1,e_2,e_3|x,y,z)=x^2(y-z)+y^2(z-x)+z^2(x-y)=0$$ gives that two variables must be equal(?). WLOG let $y=z$.

We need to solve this problem: Minimize $x^2+2xy+3y^2$ Subject to $2x^2y^2+y^4+x^2y^4=4.$ I guess WolframAlpha can solve this optimization problem.