When replying to the problem, I ask that you make posts for solutions and submit comments, jokes, smilies, etc. separately. Please use LaTeX for posting solutions. Thanks.

Note, that we can assume $a+b+c=1$. Then $a,b,c\in (0,1)$ and our inequality appears this way: $\frac{(a+1)^2}{2a^2+(1-a)^2}+\frac{(b+1)^2}{2b^2+(1-b)^2}+\frac{(c+1)^2}{2c^2+(1-c)^2}\leqslant 8$. Let's put $f(x)=\frac{(x+1)^2}{2x^2+(1-x)^2}$ for $x\in (0,1)$. This function has the following properties: 1) Its derivative: $f'(x)=-\frac{4(x+1)(2x-1)}{(3x^2-2x+1)^2}$. We see that for $x\in (0,\frac{1}{2})$ function $f$ is increasing; for $x\in (\frac{1}{2},1)$ it is decreasing. Its maximum is reached for $x=\frac{1}{2}$ and equals $f_{max}=f(\frac{1}{2})=3$, 2) $f(\frac{1}{3})=\frac{8}{3}$, $f(\frac{1}{4})=2\frac{3}{11}<\frac{7}{3}$ and $f(\frac{1}{5})=2$, 3) By calculating the second derivative one may check that $f$ is convex in $(\frac{1}{5},\frac{1}{2})$. Now let's see what follows from above. If at least one of the numbers $a,b,c$ does not exceed $\frac{1}{5}$, our ineq. obviously holds. Assume $a,b,c>\frac{1}{5}$. If we have e.g. $a\geqslant \frac{1}{2}$, then $b,c<\frac{1}{3}$ and more - one of $b,c$, for example $b$ must obey $b\leqslant \frac{1}{4}$ - and then of course the ineq. holds. Finely, if all three numbers are s.t. $a,b,c\in (\frac{1}{5},\frac{1}{2})$ - then use Jensen's ineq. to prove the given ineq.

I'd be happy to see some more interesting solution...

See also http://www.mathlinks.ro/Forum/viewtopic.php?t=46 http://www.mathlinks.ro/Forum/viewtopic.php?t=28678 http://www.mathlinks.ro/Forum/viewtopic.php?t=43278 http://www.mathlinks.ro/Forum/viewtopic.php?t=48989 darij

MithsApprentice wrote: Let $a, b, c$ be positive real numbers. Prove that \[ \frac{(2a+b+c)^2}{2a^2+(b+c)^2} + \frac{(2b+c+a)^2}{2b^2+(c+a)^2} + \frac{(2c+a+b)^2}{2c^2+(a+b)^2} \le 8. \] $\frac{(2a+b+c)^2}{2a^2+(b+c)^2}\leq\frac{4}{3}\cdot\frac{4a+b+c}{a+b+c}\Leftrightarrow(2a-b-c)^2(5a+b+c)\geq0.$

thank you arqady your solution very nice

dieuhuynh wrote: thank you arqady your solution very nice I am glad, that it was pleasant to you.

assume a+b+c=1 (2a+b+c)^2=2a^2+(b+c)^2+2a^2+4a(b+c)=2a^2+(b+c)^2+2a(2-a) (2a+b+c)^2/(2a^2+(b+c)^2) + (2b+a+c)^2/(2b^2+(a+c)^2) + (2c+a+b)^2/(2c^2+(a+b)^2) = 3+2a(2-a)/(2a^2+(b+c)^2)+2b(2-b)/(2b^2+(a+c)^2)+2c(2-c)/(2c^2+(a+b)^2) 3((2a)^2+(b+c)^2+(b+c)^2)>=(2a+2b+2c)^2=4 2a^2+(b+c)^2>=2/3 2a(2-a)/(2a^2+(b+c)^2)+2b(2-b)/(2b^2+(a+c)^2)+2c(2-c)/(2c^2+(a+b)^2)<=3/2(4a+4b+4c-2(a^2-b^2-c^2)) =6-3(a^2+b^2+c^2) a^2+b^2+c^2>=1/3(a+b+c)^2=1/3 6-3(a^2+b^2+c^2)<=6-1=5 (2a+b+c)^2/(2a^2+(b+c)^2)+(2b+a+c)^2/(2b^2+(a+c)^2)+(2c+a+b)^2/(2c^2+(a+b)^2)<=3+5=8

paladin8 wrote:

Is it true if we similar to above as$a+b+c=3$or$1$ If it is true than I can do $a+b+c=k$ where $k$ is element of Real number Abdurashid

Yes, for any real number k we can set it to a+b+c (this follows from the fact that we can rescale the inequality). In other words, see what happens when you plug in ka, kb, and kc for a,b,c....

can you set a value to $abc$ too?, not that i'm saying you would in this problem.

if you could set up the same scaling type of argument on that term

I have come up with this less beautiful solution: Suppose $8-\sum \frac{(2a+b+c)^{2}}{2a^{2}+(b+c)^{2}}=\frac{A}{B}$ We know B>0. A=$\sum_{sym}(4a^{6}+4a^{5}b+a^{4}b^{2}+5a^{4}bc+5a^{3}b^{3}-26a^{3}b^{2}c+7a^{2}b^{2}c^{2})$ We must prove that A>0. Since $4a^{6}+b^{6}+c^{6}\ge 6a^{4}bc$ and $3a^{5}b+3a^{5}c+b^{5}a+c^{5}a \ge 8a^{4}bc$ we have: $\sum_{sym}6a^{6}\ge \sum_{sym}6a^{4}bc$ and $\sum_{sym}8a^{5}\ge \sum_{sym}8a^{4}bc$ Therefore, $\sum_{sym}(4a^{6}+4a^{5}b+5a^{4}bc) \ge \sum_{sym}13a^{4}bc$ From AM-GM, we have $a^{4}b^{2}+b^{4}c^{2}+c^{4}a^{2}\ge 3a^{2}b^{2}c^{2}$ and $a^{3}b^{3}+b^{3}c^{3}+a^{3}c^{3}\ge 3a^{2}b^{2}c^{2}$ So $\sum_{sym}(a^{4}b^{2}+5a^{3}b^{3}) \ge \sum_{sym}6a^{2}b^{2}c^{2}$ Or $\sum_{sym}(a^{4}b^{2}+5a^{3}b^{3}+7a^{2}b^{2}c^{2}) \ge \sum_{sym}13a^{2}b^{2}c^{2}$. So our original sum is $\sum_{sym}(13a^{4}bc-26a^{3}b^{2}c+13a^{2}b^{2}c^{2}) \ge 13 abc \sum_{sym}(a^{3}-2a^{2}b+abc) \ge 0$ which is Schur.

arqady wrote: dieuhuynh wrote: thank you arqady your solution very nice I am glad, that it was pleasant to you. It's nice but a little overused the first two solutions in darij's links use the exact same... see nttu's post in the second and the last post of the first \[\frac43\frac{(a+1)(2a^{2}+(3-a)^{2})-(3+a)^{2}}{(a-1)^{2}}=4a+3\]

Note that we can obtain that linear bound by using calculus to obtain the tangent line at the point where equality occurs.

someone say ((2a+b+c)^2)/(2a^2+(b+c)^2)<=4(4a+b+c)/3(a+b+c) why? And why it yield the following...

greentreeroad wrote: someone say ((2a+b+c)^2)/(2a^2+(b+c)^2)<=4(4a+b+c)/3(a+b+c) why? Because $ \frac{(2a+b+c)^2}{2a^2+(b+c)^2}\leq\frac{4}{3}\cdot\frac{4a+b+c}{a+b+c}\Leftrightarrow(2a-b-c)^2(5a+b+c)\geq0.$ greentreeroad wrote: And why it yield the following... Because $ \sum_{cyc}\frac{(2a+b+c)^2}{2a^2+(b+c)^2}\leq\sum_{cyc}\left(\frac{4}{3}\cdot\frac{4a+b+c}{a+b+c}\right)=8.$

sorry, this is just a sketch because I wrote this in a few minutes in a private message to myself. $\sum_{cyc}\frac{(3+a)^2}{2a^2+(3-a)^2}\le 8$ $\frac{(3+a)^2}{2a^2+(3-a)^2}\le h(1)+(x-1)h'(1)=8/3+4(x-1)/3=4\frac{x+1}{3}$ f(x)=(3+a)^2, g(x)=3a^2-6a+9, h(x)=f(x)/g(x); h'(1)=4/3 Summing cyclically, LHS\leq4/3(6)=8, we can prove this is actually true by factoring

$$ \frac{(2a + b + c)^2}{2a^2 + (b + c)^2} + \frac{(2b + c + a)^2}{2b^2 + (c + a)^2} + \frac{(2c + a + b)^2}{2c^2 + (a + b)^2} \le 8.$$By homogenity say a+b+c=1 INEQUALITY becomes $$ \sum{\frac{(a+1)^2}{2a^2 + (1-a)^2}} \le 8.$$$$\sum{\frac{a^2+2a+1}{3a^2-2a+1}} =\sum{\frac{(a+1)^2}{2a^2 + (1-a)^2}}$$$$\sum{\frac{a^2+2a+1}{3a^2-2a+1}}-\frac{1}{3} \le 7$$$$\sum{\frac{8a+2}{3a^2-2a+1}} \le 21.$$By AM-GM $$\sum{\frac{8a+2}{3a^2-2a+1}} \le \sum{\frac{8a+1}{\frac{2}{3}}}$$$$\sum{\frac{8a+2}{\frac{2}{3}}}=\frac{8(a+b+c)+6}{\frac{2}{3}}$$$$\frac{8(a+b+c)+6}{\frac{2}{3}}=21$$Aoxz

3-variety inequalities is no more important but yep

MithsApprentice wrote: Let $ a$, $ b$, $ c$ be positive real numbers. Prove that \[ \dfrac{(2a + b + c)^2}{2a^2 + (b + c)^2} + \dfrac{(2b + c + a)^2}{2b^2 + (c + a)^2} + \dfrac{(2c + a + b)^2}{2c^2 + (a + b)^2} \le 8. \] Let $a,\,b,\,c$ are real numbers such that $a+b+c \ne 0$ and $ab+bc+ca \geqslant 0.$ Prove that $$\frac{(2a+b+c)^2}{2a^2+(b+c)^2}+\frac{(2b+c+a)^2}{2b^2+(c+a)^2}+\frac{(2c+a+b)^2}{2c^2+(a+b)^2} \geqslant 4.$$Let $ a$, $ b$, $ c$ be positive real numbers. Prove that $$ \dfrac{(2a + b + c)^2}{2a^2 + (b + c)^2} + \dfrac{(2b + c + a)^2}{2b^2 + (c + a)^2} + \dfrac{(2c + a + b)^2}{2c^2 + (a + b)^2} \ge 12-\dfrac{(a+b)(b+c)(c+a)}{2abc}$$h h h h

Have made a generalization for this one

Generalization 1 Let $a,b,c$ be positive reels. Then prove that $$\sum_{cyc}{\dfrac{\left((k+1)a+kb+kc\right)^2}{(k+1)a^2+kb^2+kc^2}}\leq 6k+\dfrac{3}{2k+1}+1$$

USAMO 2003 #5- Generalization

Generalization 2 Let $a_{1},a_{2},\cdots,a_{n}$ be positive reels. Then prove that $$\sum_{cyc}{\dfrac{\left((k+1)a_{1}+ka_{2}+ka_{3}\right)^2}{(k+1)a_{1}^2+ka_{2}^2+ka_{3}^2}}\geq n(2k-1)+\dfrac{n}{2k+1}+4$$

As the fraction’s terms have the same powers, we can set $a+b+c=3$, so that the interval we’re working with is $(0, 3)$. It suffices to prove \[\sum_{cyc}\frac{(3+a)^2}{2a^2+(3-a)^2}\le 8.\]Note that $f(x)=\tfrac{(3+x)^2}{2x^2+(3-x)^2}$ is not concave (as in, 2nd derivative isn’t obviously negative), which calls for the tangent line trick (showing that $f(x)\le f(1) + f'(1)(x-1)$). Using some calculus and simplification, this turns into \[\frac{x^2+6x+9}{x^2-2x+3}\le 4x+4\Longleftrightarrow 0\le (4x+3)(x-1)^2\]which is clearly true. Since $3f(1)=8$, this proves the result. $\square$

Wait this is very natural. $a+b+c=3$ since there's an absurd amount of symmetry. \[\frac{a^2+6a+9}{2a^2+(3-a)^2}\]and apply Tangent line trick to kill problem.

Let $a+b+c=3$ This becomes $\sum_{cyc} \frac{(a+3)^2}{2a^2+(3-a)^2}$ Let $f(a)=\frac{(a+3)^2}{2a^2+(3-a)^2}$ Then, $\frac{4x+4}{3} \ge f(x)$ because $\frac{4x+4}{3}-f(x) \ge 0$ because take the derivative to get that $0$ and $1$ are the local maximum/minimums, but at $0$ and $1$, the difference is positive, so the difference is always positive. So $f(a)+f(b)+f(c) \le \frac{4(a+b+c)}{3}+3*\frac{4}{3} = 4+4 = 8$

MithsApprentice wrote: Let $ a$, $ b$, $ c$ be positive real numbers. Prove that \[ \dfrac{(2a + b + c)^2}{2a^2 + (b + c)^2} + \dfrac{(2b + c + a)^2}{2b^2 + (c + a)^2} + \dfrac{(2c + a + b)^2}{2c^2 + (a + b)^2} \le 8. \] Note that: $$3-\frac{(2a+b+c)^2}{2a^2+(b+c)^2}=\frac{2(a-b-c)^2}{2a^2+(b+c)^2}$$So we just need to prove: $$\sum \frac{(b+c-a)^2}{2a^2+(b+c)^2} \ge \frac{1}{2}$$But: $$\sum \frac{(b+c-a)^2}{2a^2+(b+c)^2} \ge \frac{\sum (b+c-a)^2}{2(a^2+b^2+c^2)}$$and: $$\sum (b+c-a)^2 \ge \sum a^2$$Equality only holds iff $a=b=c>0$

WLOG assume $a+b+c=3$, then the inequality is equivalent to $$\sum_{cyc} \frac{(2a+3-a)^2}{2a^2+(3-a)^2} = \sum_{cyc} \frac{a^2+6a+9}{3a^2-6a+9} \leq 8 \iff \sum_{cyc} \frac{4a+3}{a^2-2a+3} \leq \frac{21}{2}.$$Using the tangent line trick, it suffices to show that $$\frac{4x+3}{x^2-2x+3} \leq \frac{7}{2} + 2(x-1)$$for all $x \in (0,3)$. But this is equivalent to $$\frac{1}{x^2-2x+3} \leq \frac{1}{2} \iff (x-1)^2 \geq 0,$$which is clearly true. $\square$

Note that the inequality \[\frac{(1+a)^2}{2a^2+(1-a)^2} \leq 4a + \frac 43 \iff \frac 13(3x-1)^2(4x+1) \ge 0\] holds over the positives. Since our desired inequality is homogenous, we can substitute $a+b+c=1$ to find \[\sum \frac{(2a+b+c)^2}{2a^2+(b+c)^2} = \sum \frac{(1+a)^2}{2a^2+(1-a)^2} \leq \sum \left(4a + \frac 43\right) = 8. \quad \blacksquare\]

Since the inequality is homogeneous, we can assume that $a+b+c=3$. \[\sum{\frac{(a+3)^2}{3a^2-6a+9}}\overset{?}{\leq} 8\iff \sum{\frac{a^2+6a+9}{a^2-2a+3}}\overset{?}{\leq} 24\]\[\sum{\frac{a^2+6a+9}{a^2-2a+3}}=3+\sum{\frac{8a+6}{a^2-2a+3}}\leq 3+4\sum{a}+9=24\]As desired.$\blacksquare$