If $a,b,c\ge 0: ab+bc+ca=1,$ prove $\frac{1}{\sqrt{a+1}}+\frac{1}{\sqrt{b+1}}+\frac{1}{\sqrt{c+1}}\le 1+\sqrt{2}.$

Question

Problem. Let $a,b,c\ge 0: ab+bc+ca=1.$ Prove that$$\frac{1}{\sqrt{a+1}}+\frac{1}{\sqrt{b+1}}+\frac{1}{\sqrt{c+1}}\le 1+\sqrt{2}.$$Equality holds at $(0,1,1).$

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I've tried to use Jichen lemma .

Firstly, we rewrite the OP as$$1+\sqrt{\frac{1}{2}}+\sqrt{\frac{1}{2}}\ge \frac{1}{\sqrt{a+1}}+\frac{1}{\sqrt{b+1}}+\frac{1}{\sqrt{c+1}}.$$ By using the lemma, we will prove three following inequalities$$2\ge \frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}.$$ $$\frac{5}{4}\ge \sum_{cyc}\frac{1}{a+1}\frac{1}{b+1}$$ $$\color{red}{\frac{1}{4}\ge \frac{1}{a+1}\frac{1}{b+1}\frac{1}{c+1}.}$$ The last inequality is wrong already which says that Jichen lemma is not appropriate.

I hope to see some good ideas. Thank you.

Answer 1

Proof.

We have, for all $x \ge 0$ $$\frac{1}{\sqrt{x + 1}} \le \frac{2\sqrt 2 - 1}{7} + \frac{24 + 22\sqrt 2}{49x + 35 + 28\sqrt 2}. \tag{1}$$ (Note: We have $\mathrm{RHS}^2 - \mathrm{LHS}^2 = \frac{(9 - 4\sqrt 2)x(x-1)^2}{(7x + 5 + 4\sqrt 2)^2(x + 1)}\ge 0$.)

Using (1), it suffices to prove that $$3 \cdot \frac{2\sqrt 2 - 1}{7} + \sum_{\mathrm{cyc}} \frac{24 + 22\sqrt 2}{49a + 35 + 28\sqrt 2} \le 1 + \sqrt 2$$ or $$\sum_{\mathrm{cyc}} \frac{2}{7a + 5 + 4\sqrt 2} \le \sqrt 2 - 1$$ or (after clearing the denominators) $$(21 + 14\sqrt 2)(ab + bc + ca) + (5 + 4\sqrt 2)(a + b + c) + 49abc - 31 - 22\sqrt 2 \ge 0. \tag{2}$$

Using $(a + b + c)^2 \ge 3(ab + bc + ca)$, we have $a + b + c \ge \sqrt 3$.

If $a + b + c \ge 2$, we have $$\mathrm{LHS}_{(2)} \ge 21 + 14\sqrt 2 + (5 + 4\sqrt 2)\cdot 2 - 31 - 22\sqrt 2 = 0.$$

If $\sqrt 3 \le a + b + c < 2$, using $abc \ge \frac{4(a + b + c)(ab + bc + ca) - (a + b + c)^3}{9}$ (degree three Schur inequality), we have \begin{align*} \mathrm{LHS}_{(2)} &\ge -\frac{49}{9}(a + b + c)^3 + (241/9 + \sqrt 2)(a + b + c) - 10 - 8\sqrt 2\\[6pt] &\ge 0. \end{align*}

We are done.

Answer 2

If $c=0$, so $ab=1$ and by C-S $$\sum_{cyc}\frac{1}{\sqrt{1+a}}=\frac{1}{\sqrt{1+a}}+\frac{1}{\sqrt{1+b}}+1\leq\sqrt{2\left(\frac{1}{1+a}+\frac{1}{1+b}\right)}+1=$$ $$=\sqrt{\frac{2(2+a+b)}{2+a+b}}+1=\sqrt2+1.$$ Now, let $f(a,b,c,\lambda)=\sum\limits_{cyc}\frac{1}{\sqrt{1+a}}+\lambda(ab+ac+bc-1)$ and let $(a,b,c)$ be an inside maximal point.

Thus, in this point should be $$\frac{\partial f}{\partial a}=\frac{\partial f}{\partial b}=\frac{\partial f}{\partial c}=0$$ or $$-\frac{1}{2\sqrt{(1+a)^3}}+\lambda(b+c)=-\frac{1}{2\sqrt{(1+b)^3}}+\lambda(a+c)=-\frac{1}{2\sqrt{(1+c)^3}}+\lambda(a+b)=0,$$ which gives $$(a+1)^3(b+c)^2=(b+1)^3(a+c)^2=(c+1)^3(a+b)^2.$$ Now, let in this point $a\neq b$ and $a\neq c$.

Thus, $$(a+1)^3(b+c)^2=(b+1)^3(a+c)^2$$ gives $$(a-b)(1-a-b+c+3c^2-3ab+3abc-c^2ab)=0$$ and $$(a+1)^3(b+c)^2=(c+1)^3(a+b)^2$$ gives $$(a-c)(1-a-c+b+3b^2-3ac+3abc-b^2ac)=0,$$ which gives $$1-a-c+b+3b^2-3ac+3abc-b^2ac=1-a-b+c+3c^2-3ab+3abc-c^2ab$$ or $$(b-c)(2+3a+3b+3c-abc)=0$$ and since by AM-GM $$abc\leq\sqrt{\left(\frac{ab+ac+bc}{3}\right)^3}=\frac{1}{3\sqrt3}<1,$$ we obtain $b=c$ and it's enough to prove our inequality for equality case of two variables.

Let $b=a$ and $c=\frac{1-a^2}{2a},$ where $0<a\leq1.$

Thus, it's enough to prove that: $$\frac{2}{\sqrt{1+a}}+\frac{1}{\sqrt{1+\frac{1-a^2}{2a}}}\leq1+\sqrt2$$ and the rest is smooth.

Can you end it now?

Answer 3

Proof

By Cauchy-Schwarz \begin{align*} &\sum_{cyc}\frac{1}{\sqrt{1+x}}\le \sqrt{\sum_{cyc}{\frac{x+1+\sqrt{2}}{x+1}}\sum_{cyc}{\frac{1}{x+1+\sqrt{2}}}}\le 1+\sqrt{2}\\&\Leftrightarrow \sum_{\mathrm{cyc}}{\left(\begin{array}{c} 4x^3y^2z+\left( \dfrac{2\sqrt{2}}{3}+3 \right) x^2y^2z^2+\\ (x+y-z)^2\left( \left( \sqrt{2}+2 \right) x^2z^2+\left( \sqrt{2}+1 \right) y^2z^2 \right) +\left( 4\sqrt{2}+6 \right) xy^4z \end{array} \right)}\\&+4\left( 5\sqrt{2}+6 \right) xyz(xy+xz+yz)^{3/2}\ge 0 \end{align*} We end proof here.

Answer 4

Source? I remember it belongs Tran Nk Trang. Here is just a sketch of author's idea.

• Firstly, we prove the following lemma.

If $x,y,z\ge 0$ such that $x^2+y^2+z^2\le 2$ and $$x^2y^2+y^2z^2+z^2x^2+(2+2\sqrt{2})x^2y^2z^2\le \dfrac{7+2\sqrt{2}}{4}$$then $x+y+z\le 1+\sqrt{2}.$

• Apply the lemma by setting $x=\dfrac{1}{\sqrt{a+1}};y=\dfrac{1}{\sqrt{b+1}};z=\dfrac{1}{\sqrt{c+1}}.$

We will prove two inequalities $$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\le 2. \tag{1}$$ $$a+b+c+5+2\sqrt{2}\le \frac{7+2\sqrt{2}}{4}(a+1)(b+1)(c+1). \tag{2}$$ Can you end it now?