Prove that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le\frac{3}{\sqrt{7}}$

Question

Let $a,b,c>0$ such that $$\dfrac{1}{a^2+2}+\dfrac{1}{b^2+2}+\dfrac{1}{c^2+2}=\dfrac{1}{3}.$$ Show that $$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\le\dfrac{3}{\sqrt{7}}.$$

My try: since $$\dfrac{1}{2+a^2}=\dfrac{1}{2}\left(1-\dfrac{a^2}{a^2+2}\right)$$ so $$\dfrac{a^2}{a^2+2}+\dfrac{b^2}{b^2+2}+\dfrac{c^2}{c^2+2}=\dfrac{7}{3}$$ we only prove

$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\le\dfrac{3}{\sqrt{7}}$

and I want use Cauchy-Schwarz inequality to prove it,But I can't works,such $$\left(\dfrac{a^2}{a^2+2}+\dfrac{b^2}{b^2+2}+\dfrac{c^2}{c^2+2}\right)(a^2+2+b^2+2+c^2+2)\ge (a+b+c)^2$$ $$(a^2+b^2+c^2+6)\ge \dfrac{3}{7}(a^2+b^2+c^2+2ab+2bc+2ac)$$ $$\Longrightarrow 4(a^2+b^2+c^2)+42\ge 6(ab+bc+ac)$$ and let $$p=a+b+c,q=ab+bc+ac,r=abc$$ so $$2p^2+21\ge 7q$$ and we only prove $$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{ab+bc+ac}{abc}=\dfrac{q}{r}\le\dfrac{3}{\sqrt{7}}$$ maybe this is not true.

But this not usefull to solve this problem .

Thank you

Answer 1

Let $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}>\frac{3}{\sqrt7}$, $\frac{1}{a}=\frac{kx}{\sqrt7}$, $\frac{1}{b}=\frac{ky}{\sqrt7}$ and $\frac{1}{c}=\frac{kz}{\sqrt7}$, where $k>0$ and $x+y+z=3$.

Hence, $k>1$ and $\frac{1}{3}=\sum\limits_{cyc}\frac{1}{a^2+2}=\sum\limits_{cyc}\frac{1}{\frac{7}{k^2x^2}+2}>\sum\limits_{cyc}\frac{1}{\frac{7}{x^2}+2}=\sum\limits_{cyc}\frac{x^2}{7+2x^2}$,

which is a contradiction because we'll prove now that $\sum\limits_{cyc}\frac{x^2}{7+2x^2}\geq\frac{1}{3}$.

Indeed, let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.

Hence, we need to prove that $$4x^2y^2x^2+\sum\limits_{cyc}(8x^2y^2+7x^2)-49\geq0$$ or $$4w^6+8(9v^4-6uw^3)u^2+7(9u^2-6v^2)u^4-49u^6\geq0$$ or $f(w^3)\geq0$, where $f(w^3)=2w^6-24u^3w^3+7u^6-21u^4v^2+36u^2v^4$.

We see that $f'(w^3)=4w^3-24u^3<0$, which says that it's enough to prove that

$f(w^3)\geq0$ for a maximal value of $w^3$, which happens for equality case of two variables.

Let $y=x$ and $z=3-2x$.

Hence, $\sum\limits_{cyc}\frac{x^2}{7+2x^2}\geq\frac{1}{3}\Leftrightarrow(x-1)^2(2x-1)^2\geq0$.

Done!

Also we can use the Vasc's LCF Theorem.

Answer 2

An analytical proof is proposed below :

Answer 3

This might not be the "proper" way, but it can be seen that both the expressions are symmetrical in all terms. And we need the maximum of the second eqn. This can happen only when all the terms contribute equally to the sum, if it were not so then, let 1/a<1/b<1/c, then we can make b and c to a, and get higher sum. So, putting it back in the first eqn. we get $ \frac{3}{a^2+2} = 1/3$, solving which we get $ a = \sqrt{7}$ or $-\sqrt{7} $. Choosing the positive root, as we need maximum values, we get $ \frac{1}{a} + \frac{1}{b}+\frac{1}{c} = \frac{3}{\sqrt{7}}$

Missed JJacquelin's answer, by a few minutes. It is more analytical, than mine.

Answer 4

By AM-GM, $1=\dfrac{3}{a^2+2}+\dfrac{3}{b^2+2}+\dfrac{3}{c^2+2} \geq \dfrac{9}{\sqrt[3]{(a^2+2)(b^2+2)(c^2+2)}}$

By AM-GM, $7 = \dfrac{3a^2}{a^2+2}+\dfrac{3b^2}{b^2+2}+\dfrac{3c^2}{c^2+2} \geq \dfrac{9\sqrt[3]{a^2b^2c^2}}{\sqrt[3]{(a^2+2)(b^2+2)(c^2+2)}}$

As everything is positive, we can safely divide the inequalities with each other.

$\dfrac{7}{1} \geq \dfrac{\dfrac{9\sqrt[3]{a^2b^2c^2}}{\sqrt[3]{(a^2+2)(b^2+2)(c^2+2)}}}{\dfrac{9}{\sqrt[3]{(a^2+2)(b^2+2)(c^2+2)}}}$

$\Rightarrow 7 \geq (abc)^{\dfrac{2}{3}}$

$\Rightarrow \dfrac{1}{\sqrt[3]{abc}} \geq \dfrac{1}{\sqrt{7}}$

By AM-GM, $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} \geq \dfrac{3}{\sqrt[3]{abc}}\geq \dfrac{3}{\sqrt{7}}$

$\therefore \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} \geq \dfrac{3}{\sqrt{7}}$

Answer 5

Alternative proof, without using uvw:

Let $x = \frac{\sqrt 7}{a}, y = \frac{\sqrt 7}{b}, z = \frac{\sqrt 7}{c}$. We have $x, y, z > 0$ and $$\frac{1}{2x^2 + 7} + \frac{1}{2y^2 + 7} + \frac{1}{2z^2 + 7} = \frac13. \tag{1}$$ We need to prove that $$x + y + z \le 3. $$

WLOG, assume that $x \le y \le z$. Let $w = \frac{x + y}{2}$.

Fact 1: $x + y \le 2$.
(The proof is given at the end.)

Fact 2: It holds that $$\frac{1}{2x^2 + 7} + \frac{1}{2y^2 + 7} \le 2 \cdot \frac{1}{2w^2 + 7}.$$ (The proof is given at the end.)

By Fact 2 and (1), we have $$2 \cdot \frac{1}{2w^2 + 7} + \frac{1}{2z^2 + 7} \ge \frac13$$ or $$\frac{2(2w^2z^2 + 4w^2 + z^2 - 7)}{3(2w^2+7)(2z^2+7)} \le 0.$$ Thus, we have $$2w^2z^2 + 4w^2 + z^2 - 7 \le 0$$ or $$2(wz - 1)^2 + (2w + z)^2 \le 9$$ which results in $$2w + z \le 3$$ that is $$x + y + z \le 3.$$

We are done.

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Proof of Fact 1:

From (1) and $x \le y \le z$, we have $$\frac{1}{2x^2 + 7} + \frac{1}{2y^2 + 7} \ge \frac29$$ or $$\frac{8x^2y^2 + 10x^2 + 10y^2 - 28}{9(2x^2+7)(2y^2+7)} \le 0$$ or $$8x^2y^2 + 10x^2 + 10y^2 - 28 \le 0.$$ Using $x^2y^2 + 1 \ge 2xy$, we have $$8(2xy - 1) + 10x^2 + 10y^2 - 28 \le 0$$ or $$16xy + 10x^2 + 10y^2 - 36 \le 0$$ or $$8(x + y)^2 + 2x^2 + 2y^2 - 36 \le 0$$ Using $2x^2 + 2y^2 \ge (x + y)^2$, we have $$9(x + y)^2 \le 36$$ or $$x + y \le 2.$$

We are done.

Proof of Fact 2:

It suffices to prove that $$\frac{1}{2x^2+7} - \frac{1}{2w^2 + 7} \le \frac{1}{2w^2+7} - \frac{1}{2y^2 + 7}$$ or $$\frac{2(w-x)(w + x)}{(2x^2+7)(2w^2+7)} \le \frac{2(y - w)(y + w)}{(2w^2+7)(2y^2+7)}$$ or (using $w - x = y - w = (y-x)/2 \ge 0$) $$\frac{w + x}{2x^2+7} \le \frac{y + w}{2y^2+7}$$ or $$\frac{(y - x)(2wx + 2wy + 2xy - 7)}{(2y^2+7)(2x^2+7)} \le 0$$ or $$2wx + 2wy + 2xy - 7 \le 0$$ or $$(x + y)^2 + 2xy - 7\le 0$$ which is true using $x + y \le 2$ (Fact 1).

We are done.