Use these inequality : $(2a +b+c)^2$ = $4(a+b)(a+c) + (b-c)^2 \geq 4 (a+b)(a+c) $ And $(ab+ac+bc)(a+b+c) \geq 9abc $ (Official solution is as same as mine !) I think It`s too easy , Opposite the geometries ISL ...

\[\frac{1}{(2a+b+c)^2}+\frac{1}{(a+2b+c)^2}+\frac{1}{(a+b+2c)^2}\leq \frac{3}{16}\] \[ \iff \sum_{cyc}\left(\frac{(16-(2a+b+c)^2}{16(2a+b+c)^2}\right) \leq 0 \] Nominator and Denominator are oppositely sorted, therefore: \[ \sum_{cyc}\left(\frac{(16-(2a+b+c)^2}{16(2a+b+c)^2}\right) \leq \frac{1}{3}\sum_{cyc}{(16-(2a+b+c)^2)}\cdot \sum_{cyc}{\frac{1}{16(2a+b+c)^2} \leq 0 }\] \[ \iff \sum_{cyc}{(16-(2a+b+c)^2)} \leq 0 \] Which is true because by Cauchy Schwarz inequality we have $3\sum_{cyc}{(2a+b+c)^2} \geq 16(a+b+c)^2 \geq 3\cdot 48 \iff (a+b+c)^2\geq 9$. But this is true by Cauchy Schwarz because: \[(a+b+c)^2 = \left(\frac{1}{a} +\frac{1}{b} +\frac{1}{c}\right)(a+b+c) \geq 9 \]

Using the condition and AM-HM inequality, we have that $a+b+c+\frac{1}{a}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{a} \geq \frac{16}{2a+b+c}.$ So $\sum \limits_{cyc} \frac{16}{(2a+b+c)^2} \leq \sum \limits_{cyc} \frac{a+b+c+\frac{1}{a}}{(2a+b+c)^2}=3+\sum \limits_{cyc} \frac{\frac{1}{a}-a}{2a+b+c}$. Therefore it's enough to prove that $\sum \limits_{cyc} \frac{a-\frac{1}{a}}{2a+b+c} \leq 0 (*)$. It's obviously that $a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq \frac{9}{a+b+c}$, so $a+b+c \geq 3 \eqno (**) .$ WLOG, suppose that $a \geq b \geq c$. Consider two possible cases: 1st case: $a+b \geq a+c \geq c+a \geq 1.$ In this case it's easy to see that $a-\frac{1}{a} \geq b-\frac{1}{b} \geq c-\frac{1}{c} $ and $\frac{1}{2a+b+c} \leq \frac{1}{a+2b+c} \leq \frac{1}{a+b+2c}.$ So by Chebyshev's Inequality we have that \[LHS(*) \leq \frac{1}{3}\left(a-\frac{1}{a} +b-\frac{1}{b} +c-\frac{1}{c} \right)\left(\frac{1}{2a+b+c} + \frac{1}{a+2b+c} + \frac{1}{a+b+2c}\right)=\left[a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right]=0,\] as desired. 2nd case: $b+c<1.$ Then $b, c<1 $, $b-\frac{1}{b}<0$ and $c-\frac{1}{c}<0$. From $(*)$ we have that $a>2$, so $a-\frac{1}{a}>0$. Therefore \[ \frac{a-\frac{1}{a}}{2a+b+c}+\frac{b-\frac{1}{b}}{a+2b+c}+\frac{c-\frac{1}{c}}{a+b+2c} \leq \frac{a-\frac{1}{a}}{2a+b+c}+\frac{b-\frac{1}{b}}{a+2b+c}+\frac{c-\frac{1}{c}}{a+2b+c} = \frac{a-\frac{1}{a}}{2a+b+c}+\frac{b-\frac{1}{b}+c-\frac{1}{c}}{a+2b+c}=\left[b+c- \frac{1}{b}- \frac{1}{c}=\frac{1}{a}-a\right] = \left(a-\frac{1}{a}\right)\left( \frac{1}{2a+b+c} -\frac {1}{a+2b+c} \right). \] But $\left(a-\frac{1}{a}\right)\left( \frac{1}{2a+b+c} -\frac {1}{a+2b+c} \right) < 0,$ because $a-\frac{1}{a}>0$ and $\frac{1}{2a+b+c} -\frac {1}{a+2b+c} <0$. So we have done.

Homogenizing and using AM-GM on $2a$ and $b+c$ we get \begin{align*} \sum\frac{abc(a+b+c)}{(2a+b+c)^2} \le \frac{1}{8}\sum bc\left(1+\frac{a}{b+c}\right) &= \frac{1}{8}\sum bc + \frac{1}{8}abc\sum\frac{1}{b+c}\\ &\le \frac{1}{8}\sum bc + \frac{1}{16}abc\sum\frac{1}{a} = \frac{3}{16}\sum bc, \end{align*}where the last part follows from \[\sum\left(\frac{1}{b}+\frac{1}{c}\right) \ge \sum\left(\frac{4}{b+c}\right).\]

Does this work? Let $S=a+b+c=\frac 1a+\frac 1b+\frac 1c$. Thus, the inequality is $\sum\frac{1}{(S+a)^2}\le \frac{3}{16}$. Since $\frac{1}{(S+x)^2}$ is concave, $\sum\frac{1}{(S+a)^2}\le \frac{3}{(4S/3)^2}\le\frac{3}{16}$. The latter inequality is true for $(a+b+c)(\frac 1a+\frac 1b+\frac 1c)=S^2\ge 9$.

GoldenFrog1618 wrote: Does this work? ... Since $\frac{1}{(S+x)^2}$ is concave, ... $\frac{1}{(S+x)^2}$ is convex, so this doesn't work.

Let $(x, y, z)=\left(\frac{3a}{a+b+c},\frac{3b}{a+b+c},\frac{3c}{a+b+c} \right)$. Then we have by weighted AM-GM \[ 4x \leq x^4 + 3 \\ 15x \leq 5x^3 + 10 \\ 10x \leq 5x^2 + 5. \] Adding them up, we get \[32x \leq (x+3)^2(x(x-1)+2)\\ \Rightarrow 32x - 2(x+3)^2 \leq (x+3)^2x(x-1)\\ \Rightarrow \frac{1}{(x+3)^2} - \frac{1}{16x} \leq \frac{1}{32} x - \frac{1}{32}\] and similarly for $y, z$. Adding cyclically, we find \[ \frac{1}{(x+3)^2} + \frac{1}{(y+3)^2} + \frac{1}{(z+3)^2} \leq \frac{1}{32} (x+y+z) - \frac{3}{32}+\frac{1}{16}\left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right).\] Also, $x+y+z = 3$, so we have \[(x+y+z) \sum_{\text{cyc}} \frac{1}{(2x+y+z)^2} \leq \frac{3}{16}\sum_{\text{cyc}} \frac{1}{x},\] which is homogeneous in $x, y, z$. Thus, we can scale the variables to $a, b, c$ and get \[(a+b+c) \sum_{\text{cyc}} \frac{1}{(2a+b+c)^2} \leq \frac{3}{16}\sum_{\text{cyc}} \frac{1}{a},\] which, after applying the condition, becomes \[ \frac{1}{(2a+b+c)^{2}}+\frac{1}{(a+2b+c)^{2}}+\frac{1}{(a+b+2c)^{2}}\leq\frac{3}{16}.\]

Oh! I've just seen that problem, it's really nice, but very easy for IMO. we have $3(ab+bc+ca)=3abc(a+b+c)\leq (ab+bc+ca)^2$ then $ab+bc+ca\geq 3$ then $\sum \frac{16}{(a+b+2c)^2}\leq \frac{8\sum a}{\Pi (a+b)}\leq \frac{9}{ab+bc+ca}\leq 3$ Done. I've just used the fact that: $(a+b+2c)^2=(a+c+b+c)^2\geq 4(a+c)(b+c)$ and that $(a+b)(b+c)(c+a)\geq \frac{8}{9}(a+b+c)(ab+bc+ca)$

First, we eliminate the constraint by homogenizing the inequality. We wish to show that for all positive $a,b,c$, $\frac{1}{(2a+b+c)^2} + \frac{1}{(a+2b+c)^2} + \frac{1}{(a+b+2c)^2} \leq \frac{3\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)}{16(a+b+c)}$. This inequality is homogenous, so we may assume without loss of generality that $a+b+c = 3$. Hence, if we let $f(x) = \frac{1}{(3+x)^2} - \frac{1}{16x}$, it is sufficient to show that $f(a) + f(b) + f(c) \leq 0$ when $a+b+c = 3$. $\frac{1}{(x+3)^2} - \frac{1}{16x} \leq \frac{x-1}{32}$ is equivalent to $\frac{(x-1)^2(x^2 + 7x + 18)}{32x(x+3)^2} \geq 0$, which is always true. Hence, $f(a) + f(b) + f(c) \leq \frac{a+b+c-3}{32} = 0$, as desired. Note: If I'm not mistaken, Jensen's inequality could also be used to show that $f(a) + f(b) + f(c) \leq 3f \left (\frac{a+b+c}{3} \right) = 0$.

Also the stronger inequality holds with the same condition: \[\frac{1}{(2a+b+c)^4}+\frac{1}{(a+2b+c)^4}+\frac{1}{(a+b+2c)^4} \le \frac{3}{2^8}\]

I think with the same condition we can prove the more general inequality for $n \in \mathbb{N}$. \[ \frac{1}{(2a+b+c)^n} +\frac{1}{(a+2b+c)^n} +\frac{1}{(a+b+2c)^n} \leq \frac{3}{4^n}\]

enndb0x wrote: I think with the same condition we can prove the more general inequality for $n \in \mathbb{N}$. \[ \frac{1}{(2a+b+c)^n} +\frac{1}{(a+2b+c)^n} +\frac{1}{(a+b+2c)^n} \leq \frac{3}{4^n}\] Nobody prove its?

mahanmath wrote: Use these inequality : $(2a +b+c)^2$ = $4(a+b)(a+c) + (b-c)^2 \geq 4 (a+b)(a+c) $ And $(ab+ac+bc)(a+b+c) \geq 9abc $ (Official solution is as same as mine !) I think It`s too easy , Opposite the geometries ISL ... If we work out with your method, we have $LL\le \frac{2(a+b+c)}{4(a+b)(b+c)(a+c)}$ Which comes similar with $8(a+b+c)\le 3(a+b)(b+c)(a+c)$ But why follows this of $(ab+ac+bc)(a+b+c) \geq 9abc ?$ To answer about last post; try $a=2/3,b=1,c=1.5$ and for big n: $(\frac{4}{2a+b+c})^n>3$

$ (ab+ac+bc)(a+b+c)\geq 9abc $ implies $9(a+b)(b+c)(c+a) \geq 8(a+b+c)(ab+bc+ca) $ (#) And the condition is $ab+bc+ca=abc(a+b+c)$ so use (#) the main inequality boils down to $(ab+bc+ca)^2 \geq 3abc(a+b+c)$ Which is wellknown .

At first, I understand your way, Mahanmath. We can maybe try to find the most big number $n$ it satisfy, any idea for $n=4$ already? Pirshtuk wrote: Using the condition and AM-HM inequality, we have that $a+b+c+\frac{1}{a}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{a} \geq \frac{16}{2a+b+c}.$ So $\sum \limits_{cyc} \frac{16}{(2a+b+c)^2} \leq \sum \limits_{cyc} \frac{a+b+c+\frac{1}{a}}{(2a+b+c)^2}=3+\sum \limits_{cyc} \frac{\frac{1}{a}-a}{2a+b+c}$. Therefore it's enough to prove that $\sum \limits_{cyc} \frac{a-\frac{1}{a}}{2a+b+c} \leq 0 (*)$. It's obviously that $a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq \frac{9}{a+b+c}$, so $a+b+c \geq 3 \eqno (**) .$ WLOG, suppose that $a \geq b \geq c$. Consider two possible cases: 1st case: $a+b \geq a+c \geq c+a \geq 1.$ In this case it's easy to see that $a-\frac{1}{a} \geq b-\frac{1}{b} \geq c-\frac{1}{c} $ and $\frac{1}{2a+b+c} \leq \frac{1}{a+2b+c} \leq \frac{1}{a+b+2c}.$ So by Chebyshev's Inequality we have that \[LHS(*) \leq \frac{1}{3}\left(a-\frac{1}{a} +b-\frac{1}{b} +c-\frac{1}{c} \right)\left(\frac{1}{2a+b+c} + \frac{1}{a+2b+c} + \frac{1}{a+b+2c}\right)=\left[a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right]=0,\] as desired. 2nd case: $b+c<1.$ Then $b, c<1 $, $b-\frac{1}{b}<0$ and $c-\frac{1}{c}<0$. From $(*)$ we have that $a>2$, so $a-\frac{1}{a}>0$. Therefore \[ \frac{a-\frac{1}{a}}{2a+b+c}+\frac{b-\frac{1}{b}}{a+2b+c}+\frac{c-\frac{1}{c}}{a+b+2c} \leq \frac{a-\frac{1}{a}}{2a+b+c}+\frac{b-\frac{1}{b}}{a+2b+c}+\frac{c-\frac{1}{c}}{a+2b+c} = \frac{a-\frac{1}{a}}{2a+b+c}+\frac{b-\frac{1}{b}+c-\frac{1}{c}}{a+2b+c}=\left[b+c- \frac{1}{b}- \frac{1}{c}=\frac{1}{a}-a\right] = \left(a-\frac{1}{a}\right)\left( \frac{1}{2a+b+c} -\frac {1}{a+2b+c} \right). \] But $\left(a-\frac{1}{a}\right)\left( \frac{1}{2a+b+c} -\frac {1}{a+2b+c} \right) < 0,$ because $a-\frac{1}{a}>0$ and $\frac{1}{2a+b+c} -\frac {1}{a+2b+c} <0$. So we have done. In this one there are some mistakes: He said $\sum \limits_{cyc} \frac{a-\frac{1}{a}}{2a+b+c} \leq 0 (*)$ but it have to be in reverse order for $3+\sum \limits_{cyc} \frac{\frac{1}{a}-a}{2a+b+c} \leq 3$ In his first part he made the mistake ; Chebychev give $LHS(*) \geq \frac{1}{3}\left(a-\frac{1}{a} +b-\frac{1}{b} +c-\frac{1}{c} \right)\left(\frac{1}{2a+b+c} + \frac{1}{a+2b+c} + \frac{1}{a+b+2c}\right)$ and hence it is proven, but in the second part he proved the wrong thing, isn't it.

Is it true that $\frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{c+a} \le \frac{3}{2}$ if $a, b, c > 0$ and $a + b + c = \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$?

CDP100 wrote: Is it true that $\frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{c+a} \le \frac{3}{2}$ if $a, b, c > 0$ and $a + b + c = \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$? No, $a=b=0.25$ and $c$ has a solution.

first,we can make this ineq a homogenuous one,then we can assume that $a+b+c=1$.by applying LCF-RCF theorem it's trivial then.after all,we can expand it and use SOS.

April wrote: Let $a$, $b$, $c$ be positive real numbers such that $\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} = a+b+c$. Prove that: \[\frac{1}{(2a+b+c)^2}+\frac{1}{(a+2b+c)^2}+\frac{1}{(a+b+2c)^2}\leq \frac{3}{16}\] Proposed by Juhan Aru, Estonia By the AM-HM inequality,\[(a+b+c)^2\ge9\] and by the Cauchy Schwarz inequality \[\sum a^2\ge 3\]. By Titu's lemma or Cauchy Schwarz,$\sum_{cyc} \frac{1}{(2a+b+c)^2} \leq \frac{9}{\sum{(2a+b+c)^2}}$.($\dots \boxed{1}$) \[\sum(2a+b+c)^2=\sum_{cyc}a^2+2\sum a(a+b+c) + \sum_{cyc} (a+b+c)^2\] \[=\sum_{\cyc}a^2+2(a+b+c)^2 + 3(a+b+c)^2\ge 3+2\cdot 9+3\cdot 9\ge 48\]..Putting this in the inequality $\boxed{1}$, we get what we wanted to prove.

April wrote: Let $a$, $b$, $c$ be positive real numbers such that $\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} = a+b+c$. Prove that: \[\frac{1}{(2a+b+c)^2}+\frac{1}{(a+2b+c)^2}+\frac{1}{(a+b+2c)^2}\leq \frac{3}{16}.\]Proposed by Juhan Aru, Estonia Usually try to homogenize inequalities, so let’s try that here. We want to prove $\sum_{cyc}\frac{a+b+c}{(2a+b+c)^2}\leq \frac{3}{16}\sum_{cyc}\frac{1}{a}$. Now, let a+b+c=3 to quickly rid ourselves of the 3. It suffices, then, to prove that $\sum_{cyc}\frac{16a-(a+3)^2}{16a(a+3)^2}\leq 0\iff\sum_{cyc}\frac{10a-a^2-9}{a^2+6a+9}\leq \sum_{cyc}\frac{a-1}{2}=0$, where the last step is because we can expand simply and note the preserved a=b=c=1 equality case (so there must be a factor of a-1, and we see it is a double root by easy factoring) that gives the true $(a-1)^2(a^2+7a+18) \geq 0$. I dislike these kinds of problems, because at some point there is no motivation to have an intermediate step, other than lucky guess (to be specific, that (a-1)/2). Luckily, it isn’t too out of reach, since we want something that is equal to 0 cyclically, which relates to a-1 (an equality case), and then multiply by some constant (1/2). I had to use WolframAlpha along with a friend to test a bunch of values, but since this was untimed it worked. Edit: Actually, I can apply the tangent line trick to get this conjecture much, much easier, and I just learned how to take derivatives.

This homogenizes as \[ \frac{a+b+c}{(2a+b+c)^2} + \frac{a+b+c}{(a+2b+c)^2} + \frac{a+b+c}{(a+b+2c)^2} \le \frac{3}{16a} + \frac{3}{16b} + \frac{3}{16c} \]We can homogenize $a+b+c=3$ to get \[ \frac{3}{(a+3)^2} + \frac{3}{(3+b)^2} + \frac{3}{(3+c)^2} \le \frac{3}{16a} + \frac{3}{16b} + \frac{3}{16c} \]If we let $f(x) = \frac{3}{(x+3)^2} - \frac{3}{16x}$, this becomes \[ f(a) + f(b) + f(c) \le 0 \]Note that $f(1) = 0$, so if we show concavity on $[0, 3]$ the result holds by Jensen's. This follows since \[ f''(x) = \frac{18}{(x+3)^4} - \frac{3}{8x^3} \]is strictly less than $0$.

Notice that if we multiply the $RHS$ by $\frac{\sum_{cyc}\frac{1}{a}}{\sum_{cyc}a}$ we obtain that the inequality is homogeneous, so $WLOG$ let $a+b+c=3$ With this new condition the inequality transforms into $\sum_{cyc}\frac{1}{(a+3)^2}\le\frac{3}{16}=\frac{1}{16}\left(\sum_{cyc}\frac{1}{a}\right)=\sum_{cyc}\frac{1}{16a}$ Thus the inequality boils down to $\sum_{cyc}\frac{1}{(a+3)^2}-\frac{1}{16a}\le0$ Furthermore let $f(x)=\frac{1}{(x+3)^2}-\frac{1}{16x}, f'(x)=\frac{1}{16x^2}-\frac{2}{(x+3)^2}\text{ furthermore }f''(x)=\frac{6}{(x+3)^4}-\frac{1}{8x^3}<0,\text{ for }x\in(0,3)$ Thus the function is concave for $x\in(0,3)$ Therefore $\sum_{cyc}\frac{1}{(a+3)^2}-\frac{1}{16a}=\sum_{cyc}f(a)\overset{\text{Jensen}}{\le}3f\left(\frac{\sum_{cyc}a}{3}\right)=3f(1)=0\text{ thus }\sum_{cyc}\frac{1}{(a+3)^2}-\frac{1}{16a}\le0$ $\blacksquare$. 100th post

Multiplying the RHS of the equation by $\frac{\tfrac{1}{a} + \tfrac{1}{b} + \tfrac{1}{c}}{a + b + c} $ homogenizes the inequality, so we can now WLOG $a + b + c = 3$, and then it becomes \[ \frac{1}{(a+3)^2} + \frac{1}{(b+3)^2} + \frac{1}{(c+3)^2} \le \frac{1}{16a} + \frac{1}{16b} + \frac{1}{16c},\]which is equivalent to \[ \sum_{cyc} \left( \frac{(a+3)^2 - 16a}{(a+3)^2 \cdot 16a} \right) \ge 0 \] Let $f(x) = \frac{(x+3)^2 - 16x}{(x+3)^2 \cdot 16x} = \frac{1}{16x} - \frac{1}{(x+3)^2} $. Claim: $f$ is convex on the interval $(0,3)$. Proof: The second derivative of $f(x)$ is equal to $\frac{1}{8x^3} - \frac{6}{(x+3)^4}$, which is strictly greater than $0$ for $x\in (0,3)$. $\square$ Now, we get that $f(a) + f(b) + f(c) \ge f(1) = 0$ by Jensen's, as desired.

$\frac{1}{(2a+b+c)^2}+\frac{1}{(a+2b+c)^2}+\frac{1}{(a+b+2c)^2}\leq \frac{3}{16}$ $\frac{1}{(2a+b+c)^2}+\frac{1}{(a+2b+c)^2}+\frac{1}{(a+b+2c)^2}\leq \frac{3(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})}{16(a+b+c)}$ Assume that $a+b+c=3$ $\frac{1}{(3+a)^2}+\frac{1}{(3+b)^2}+\frac{1}{(3+c)^2}\leq \frac{1}{16a}+\frac{1}{16b}+\frac{1}{16c}$ $\frac{1}{(3+a)^2}+\frac{1}{(3+b)^2}+\frac{1}{(3+c)^2}-\frac{1}{16a}-\frac{1}{16b}-\frac{1}{16c}\leq 0$ If $f(x)=\frac{1}{(3+x)^2}-\frac{1}{16x}$, then this ibecomes $f(a)+f(b)+f(c) \le f(1) = 0$, which is true by Jensen since the second derivative is always negative from 0 to 3.

There was generalization 1 here

There was generalization 2 here.

let $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=a+b+c=s$ from AM-GM, $2s=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+a+b+c \geq 6$ therefore, $s \geq 6$ assume $a \geq b \geq c$, so $a \geq 1$ and $\frac{1}{(s+a)^2}+\frac{1}{(s+b)^2}+\frac{1}{(s+c)^2}\leq\frac{3}{(s+a)^2}$ we therefore have $\frac{1}{(2a+b+c)^2}+\frac{1}{(a+2b+c)^2}+\frac{1}{(a+b+2c)^2} \leq \frac{1}{(s+a)^2}+\frac{1}{(s+b)^2}+\frac{1}{(s+c)^2}\leq\frac{3}{(4)^2}=\frac{3}{16}$

calculus is for nerds (hi tangent line trick) It suffices to prove \[\sum_{\text{cyc}}\frac{1}{(2a+b+c)^2}\le \frac{3}{16}\cdot \frac{\tfrac{1}{a}+\tfrac{1}{b}+\tfrac{1}{c}}{a+b+c}.\]Write $a+b+c=3$ to yield \[\sum_{\text{cyc}}\left(\frac{1}{a}-\frac{16}{(a+3)^2}\right)\ge 0.\] Notice \[\frac{1}{a}-\frac{16}{(a+3)^2}\ge -\frac{1}{2}(a-1)\iff (a-1)^2(a^2+7a+18)\ge 0\]and summing yields the result. $\blacksquare$

Here M1 means Method 1. Red pen means intuition and blue pen means the proof without intuition.

Attachments:

This message is deleted

ehuseyinyigit wrote: Generalization 2 Let $a,b,c$ be positive reals such that $\dfrac1a+\dfrac1b+\dfrac1c=\theta \left(a+b+c\right)$. Then prove that $$\dfrac{1}{\left(\left(\lambda +\beta\right)a+\lambda b+\beta c\right)^2}+\dfrac{1}{\left(\left(\lambda +\beta\right)b+\lambda c+\beta a\right)^2}+\dfrac{1}{\left(\left(\lambda +\beta\right)c+\lambda a+\beta b\right)^2} \leq \dfrac{3\theta}{16\lambda \beta}$$ Tangent line trick, generalization 1 is a variant of this one.

Nice A2. We have $a,b,c \in \mathbb{R^{+}}$ such that $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=a+b+c$. Claim: $\sum_{cyc} \dfrac{1}{(2a+b+c)^{2}} \leq \dfrac{3}{16}$ Proof: Note that $(2a+b+c)^{2}=((a+b)+(a+c))^{2}\geq 4(a+b)(a+c)$. Thus we wish the following inequality to be true: $$\sum_{cyc} \dfrac{1}{(a+b)(a+c)} \leq \dfrac{3}{4}$$$$ \iff 8(a+b+c) \leq 3(a+b)(b+c)(c+a) $$We can homogenize the above inequality by using the condition $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=a+b+c$. Thus, Sub-Claim: $8(a+b+c)^{2} \leq 3(a+b)(b+c)(c+a)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)$ Proof: On clearing the denominators this can be rewritten as: $$ 3(a+b)(b+c)(c+a)(ab+bc+ca) \geq 8abc(a+b+c)^{2}$$$$ \iff 3(ab(a+b)+bc(b+c)+ca(c+a)+2abc) \geq 8abc(a^{2}+b^{2}+c^{2}+2ab+2bc+2ca)$$After expanding and cancelling like terms, $$\iff 3a^{3}b^{2}+3a^{3}c^{2}+3b^{3}a^{2}+3b^{3}c^{2}+3c^{3}a^{2}+3c^{3}b^{2} \geq 2a^{3}bc+2ab^{3}c+2abc^{3}+4a^{2}b^{2}c+4a^{2}bc^{2}+4ab^{2}c^{2}$$$$ \iff 3(3,2,0) \succ (3,1,1)+2(2,2,1)$$which is indeed true because $(3,2,0) \succ (3,1,1)$ and $(3,2,0) \succ (2,2,1)$ by Muirhead's Inequality. $\blacksquare$ (QED).

Bryan I don't like you_ Yes, it was. But it seems more approachable.

Trivial by Tangent-Line Trick Note that the inequality is equivalent to $$\sum_{cyc} \frac{1}{(2a+b+c)^2} \leq \frac{3}{16} \cdot \frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}{a+b+c}.$$WLOG, assume that $a+b+c=3,$ then it suffices to show that $$\sum_{cyc} \left( \frac{1}{16a}-\frac{1}{(a+3)^2} \right) \geq 0.$$However, observe that $$\frac{1}{16a}-\frac{1}{(a+3)^2} \geq -\frac{1}{32}(a-1)$$because it reduces to $$a^4+5a^3+5a^2-29a+18 \geq 0 \iff (a-1)^2(a^2+7a+18) \geq 0.$$Thus adding these up yields the desired result. QED