Suppose that a,b,c be three positive real numbers such that a+b+c=3 . Prove that : 12+a2+b2+12+b2+c2+12+c2+a2≤34
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Tags: calculus, function, three variable inequality, algebra, Inequality
12.05.2009 01:54
khashi70 wrote: Suppose that a,b,c be three positive real numbers such that a+b+c=3 . Prove that : 12+a2+b2+12+b2+c2+12+c2+a2≤34 The best inequality of this type is the following: If a,b,c are nonnegative real numbers such that a+b+c=3, then 18+5(a2+b2)+18+5(b2+c2)+18+5(c2+a2)≤16, with equality for a=b=c, and again for a13=b=c or any permutation.
12.05.2009 08:14
khashi70 wrote: Suppose that a,b,c be three positive real numbers such that a+b+c=3 . Prove that : 12+a2+b2+12+b2+c2+12+c2+a2≤34 Write the inequality as ∑a2+b2a2+b2+2≥32, or ∑(a+b)2(a+b)2+2(a+b)2a2+b2≥32. Applying the Cauchy Schwarz Inequality, we have LHS≥4(a+b+c)2∑(a+b)2+2∑(a+b)2a2+b2=36∑(a+b)2+2∑(a+b)2a2+b2. It suffices to prove that ∑(a+b)2+2∑(a+b)2a2+b2≤24. Because 12−∑(a+b)2=43(a+b+c)2−∑(a+b)2=−13∑(a−b)2, and 12−2∑(a+b)2a2+b2=2∑(a−b)2a2+b2, this inequality is equivalent to ∑(a−b)2(6a2+b2−1)≥0. Under the assumption that a≥b≥c, we see that this inequality is obviously true if a2+b2≤6. Let us consider now the case a2+b2≥6, in this case we have 1a2+b2+2≤18, and 1a2+c2+2+1b2+c2+2≤1a2+2+1b2+2≤18−b2+1b2+2≤18+12, (because 0≤b2≤94) Hence ∑1a2+b2+2≤18+18+12=34. Our proof is completed.
12.05.2009 12:11
Vasc wrote: khashi70 wrote: Suppose that a,b,c be three positive real numbers such that a+b+c=3 . Prove that : 12+a2+b2+12+b2+c2+12+c2+a2≤34 The best inequality of this type is the following: If a,b,c are nonnegative real numbers such that a+b+c=3, then 18+5(a2+b2)+18+5(b2+c2)+18+5(c2+d2)≤16, with equality for a=b=c, and again for a13=b=c or any permutation. Primitive mixing variables also can solve this one. Solution: WLOG a=max{a,b,c} f(a,b,c)=18+5(a2+b2)+18+5(b2+c2)+18+5(c2+a2) f(a,b+c2,b+c2)=28+5(a2+(b+c)24)+18+5(b+c)22 ⇒ f(a,b+c2,b+c2)−f(a,b,c)= (b−c)2.[54(16+10a2−5b2−5c2−20bc)[8+5(a2+b2)][8+5(a2+(b+c)22][8+5(a2+c2)]+52[8+5(b+c)22][8+5(b2+c2]] Note that : [8+5(a2+b2)][8+5(a2+(b+c)22][8+5(a2+c2)]≥[8+5(b+c)22][8+5(b2+c2)]2 ⇒ f(a,b+c2,b+c2−f(a,b,c)≥ ≥(b−c)2.[54(16+10a2−5b2−5c2−20bc)+52(8+5(b2+c2)[8+5(a2+b2)][8+5(a2+(b+c)22][8+5(a2+c2)]]≥ ≥(b−c)2.[54(32+10a2−5b2−5c2−20bc)+25(b2+c2)2[8+5(a2+b2)][8+5(a2+(b+c)22][8+5(a2+c2)]]≥0 Therefore, we only need to prove the inequality in the case when b=c . Now it's a very simple.
12.05.2009 19:03
Mixing Variables method is not enought for Vasc's problem... I think that. Because it only works for the case 0≤bc≤1620. But it can show you why we have the second equality. In fact we have: f(a,b,c)=RHS−LHS And f(x,3−x2,3−x2)=5(x−1)2(5x−13)26(25x2−30x+77)(5x2−30x+61)≥0
13.05.2009 08:08
khashi70 wrote: Suppose that a,b,c be three positive real numbers such that a+b+c=3 . Prove that : 12+a2+b2+12+b2+c2+12+c2+a2≤34 Suppose that a,b,c be three positive real numbers such that a+b+c=3 .find constant k is best such that: 11+k(a2+b2)+11+k(b2+c2)+11+k(c2+a2)≤31+2k i think k<23
13.05.2009 09:16
Vasc wrote: khashi70 wrote: Suppose that a,b,c be three positive real numbers such that a+b+c=3 . Prove that : 12+a2+b2+12+b2+c2+12+c2+a2≤34 The best inequality of this type is the following: If a,b,c are nonnegative real numbers such that a+b+c=3, then 18+5(a2+b2)+18+5(b2+c2)+18+5(c2+d2)≤16, with equality for a=b=c, and again for a13=b=c or any permutation. Since a+b+c=3=const. Fix a2+b2+c2=k=constant Consider function f=18+5(k−a2) . f‴ so apply n-1 EV, we only consider f(a,a,b). Now the problem is much simple and can be solved.
11.06.2009 01:10
This Iran 2009 is an old result of Vasc http://www.mathlinks.ro/Forum/viewtopic.php?t=77182
15.08.2009 10:06
khashi70 wrote: Suppose that a, b, c be three positive real numbers such that a + b + c = 3 . Prove that : \frac {1}{2 + a^{2} + b^{2}} + \frac {1}{2 + b^{2} + c^{2}} + \frac {1}{2 + c^{2} + a^{2}} \leq \frac {3}{4} Here is my solution: Lemma :With x,y,z>0,we have: (\sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x})^2 \ge 2\sqrt{3(xy+yz+zx)}+4(x+y+z) Proof . Put m=\sqrt{b+c}+\sqrt{c+a}-\sqrt{a+b};n=\sqrt{c+a}+\sqrt{a+b}-\sqrt{b+c} p=\sqrt{a+b}+\sqrt{b+c}-\sqrt{c+a},then m,n,p>0 This inequality become: mn+np+pm \ge \sqrt{3mnp(m+n+p)} Which is obvious true. Equality holds if and only if x=y=z -Now,com back the problem It is equivalent to: \frac{a^2+b^2}{2+a^2+b^2}+\frac{b^2+c^2}{2+b^2+c^2}+\frac{c^2+a^2}{2+c^2+a^2} \ge \frac{3}{2} By Cauchy-Schwarz inequality and Lemma,we have; LHS \ge \frac{(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2})^2}{6+2(a^2+b^2+c^2)} \ge \frac{2(a^2+b^2+c^2)+ab+bc+ca}{3+a^2+b^2+c^2} (because: \sqrt{3(a^2b^2+b^2c^2+c^2a^2} \ge ab+bc+ca) = \frac{\frac{3}{2}(3+a^2+b^2+c^2)}{3+a^2+b^2+c^2}=RHS (because a+b+c= 3) Then,we have Q.E.d Equality holds if and only if a=b=c=1
17.08.2009 15:47
Your proof can be write into: LHS \geq\ \frac{(\sqrt{a^{2}+b^{2}}+\sqrt{b^{2}+c^{2}}+\sqrt{c^{2}+a^{2}})^{2}}{6+2(a^{2}+b^{2}+c^{2})} = \frac{2\sum\ a^2+2\sum\ \sqrt{(a^2+b^2)(a^2+c^2)}}{6+2(a^{2}+b^{2}+c^{2})} \geq\ \frac{2\sum\ a^2+2(\sum\ a^2+\sum\ ab)}{6+2(a^2+b^2+c^2)} = \frac{3}{2}
26.10.2009 13:20
nice solution,quykhtn-a1 Here is my solution Let f(a,b,c)=\frac {1}{2 + a^{2} + b^{2}} + \frac {1}{2 + b^{2} + c^{2}} + \frac {1}{2 + c^{2} + a^{2}} => f(a,t,t)=\frac{2}{2+a^2+t^2}+\frac{1}{2+2t^2} with t=\frac{b+c}{2} We have f(a,t,t)-f(a,b,c) =(b^2+c^2-\frac{(b+c)^2}{2})(\frac{1}{(b^2+c^2+2)(2+\frac{(b+c)^2}{2})} -\frac{1}{(4+2a^2+b^2+c^2)(4+2a^2+\frac{(b+c)^2}{2})})\ge 0 It's true because 2(b^2+c^2)\ge (b+c)^2 and 4+2a^2+b^2+c^2\ge b^2+c^2+2,4+2a^2+\frac{(b+c)^2}{2}\ge 2+\frac{(b+c)^2}{2} And then,we prove that f(a,t,t)\le\frac{3}{4} \Leftrightarrow (a-1)^2(15a^2-78a+111)\ge 0 (which is obvious true)(Q.E.D) Equality holds if and only if a=b=c=1
27.02.2010 10:12
Vasc wrote: If a,b,c are nonnegative real numbers such that a + b + c = 3, then \frac 1{8 + 5(a^2 + b^2)} + \frac 1{8 + 5(b^2 + c^2)} + \frac 1{8 + 5(c^2 + d^2)}\le \frac 1{6}, with equality for a = b = c, and again for \frac a{13} = b = c or any permutation. This inequality is true for all reals a, b and c such that a + b + c = 3. Let a + b + c = 3u, ab + ac + bc = 3v^2 and abc = w^3. Since, \prod_{cyc}(a^2 + b^2) = - w^6 + A(u,v^2)w^3 + B(u,v^2) we see that the original inequality is equivalent to f(w^3)\leq0, where f is convex function. Hence, f gets a maximal value, when w^3 gets an extremal value, which happens when two numbers from \{a,b,c\} are equal. Thus, for the poof enough to check only one case: b = c = \frac {3 - a}{2}, which gives (a - 1)^2(5a - 13)^2\geq0. Done!
03.03.2010 19:38
arqady wrote: Vasc wrote: If a,b,c are nonnegative real numbers such that a + b + c = 3, then \frac 1{8 + 5(a^2 + b^2)} + \frac 1{8 + 5(b^2 + c^2)} + \frac 1{8 + 5(c^2 + d^2)}\le \frac 1{6}, with equality for a = b = c, and again for \frac a{13} = b = c or any permutation. This inequality is true for all reals a, b and c such that a + b + c = 3. Let a + b + c = 3u, ab + ac + bc = 3v^2 and abc = w^3. Since, \prod_{cyc}(a^2 + b^2) = - w^6 + A(u,v^2)w^3 + B(u,v^2) we see that the original inequality is equivalent to f(w^3)\leq0, where f is convex function. Hence, f gets a maximal value, when w^3 gets an extremal value, which happens when two numbers from \{a,b,c\} are equal. Thus, for the poof enough to check only one case: b = c = \frac {3 - a}{2}, which gives (a - 1)^2(5a - 13)^2\geq0. Done! Maybe you mean f(u,v^2,w^3)\leq 0 and f is convex?
03.03.2010 20:00
Karamata wrote: Maybe you mean f(u,v^2,w^3)\leq 0 and f is convex? Otherwise, it doesn't follow that f will attain its maximum only if w^3 is extremal f is convex function of w^3.
22.03.2010 10:41
This is also my proof, Arqady. We can generalize this inequality to n variables as follows: Let a_1,a_2,...a_n be real numbers such that a_1 + a_2 + ... + a_n = n. If r\ge \frac {n^2 - 1}{n^2 - n - 1}, then \sum \frac 1{r + a_2^2 + ... + a_n^2} \le \frac n{r + n - 1}, with equality for a_1 = ... = a_n = 1, and for a_1 = ... = a_{n - 1} = \frac 1{n^2 - n - 1} and a_n = n - \frac {n - 1}{n^2 - n - 1}.
06.08.2010 05:08
a,b,c > 0 and a+b+c=3 \frac{1}{3+a^{2}+b^{2}}+\frac{1}{3+b^{2}+c^{2}}+\frac{1}{3+c^{2}+a^{2}}\leq\frac{3}{5} the ineq becomes to \sum{\frac{a^2+b^2}{3+a^2+b^2}} \geq \frac{6}{5} and by C-S also have \left(\sum{\frac{a^2+b^2}{3+a^2+b^2}}\right)(\sum{3+a^2+b^2}) \geq (\sum{\sqrt{a^2+b^2}})^2 That means we have to prove (\sum{\sqrt{a^2+b^2}})^2 \geq \frac{6}{5} (\sum{(3+a^2+b^2)}) \sum{(a^2+b^2)}+2\sum{\sqrt{(a^2+b^2)(a^2+c^2)}} \geq \frac{54}{5} + \frac{12}{5}\sum{a^2} 8\sum{a^2} + 10\sum{ab} \geq 54 5(a+b+c)^2 + 3\sum{a^2} \geq 54 and finish the prove.
25.04.2011 22:17
arqady wrote: Vasc wrote: If a,b,c are nonnegative real numbers such that a + b + c = 3, then \frac 1{8 + 5(a^2 + b^2)} + \frac 1{8 + 5(b^2 + c^2)} + \frac 1{8 + 5(c^2 + d^2)}\le \frac 1{6}, with equality for a = b = c, and again for \frac a{13} = b = c or any permutation. ....Hence, f gets a maximal value, when w^3 gets an extremal value, which happens when two numbers from \{a,b,c\} are equal. Dear Arqady, according to the uvw method, don't you need to check also the case a=0? Thank you, best regards, Arrigo-Sacchi
26.04.2011 01:14
Arrigo-Sacchi wrote: arqady wrote: Vasc wrote: If a,b,c are nonnegative real numbers such that a + b + c = 3, then \frac 1{8 + 5(a^2 + b^2)} + \frac 1{8 + 5(b^2 + c^2)} + \frac 1{8 + 5(c^2 + d^2)}\le \frac 1{6}, with equality for a = b = c, and again for \frac a{13} = b = c or any permutation. ....Hence, f gets a maximal value, when w^3 gets an extremal value, which happens when two numbers from \{a,b,c\} are equal. Dear Arqady, according to the uvw method, don't you need to check also the case a=0? Thank you, best regards, Arrigo-Sacchi No! Because this inequality is true for all reals a, b and c such that a+b+c=3. See my full post: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1790287#p1790287
29.04.2011 11:09
Vasc wrote: The best inequality of this type is the following: If a,b,c are real numbers such that a + b + c = 3, then \frac 1{8 + 5(b^2 + c^2)} + \frac 1{8 + 5(c^2 + a^2)} + \frac 1{8 + 5(a^2 + b^2)}\le \frac 1{6}, with equality for a = b = c, and again for \frac a{13} = b = c or any permutation. \frac{1}{6}-\sum{\frac{1}{8+5b^2+5c^2}} =\frac{125}{27}\prod{\frac{(b-c)^2}{8+5b^2+5c^2}}+\frac{5}{54}\sum{\frac{(1-5a)^2(b-c)^2}{(8+5c^2+5a^2)(8+5a^2+5b^2)}}\geq0. http://www.artofproblemsolving.com/Forum/viewtopic.php?t=359320
16.11.2016 12:27
Suppose that a, b, c be three nonnegative real numbers such that a+b+c=2 . Prove that : \frac{1}{1+a^{2}+b^{2}}+\frac{1}{1+b^{2}+c^{2}}+\frac{1}{1+c^{2}+a^{2}} \geq \frac{4}{3}Where?
17.12.2016 04:16
Suppose that a, b, c be three positive real numbers such that ab+bc+ca=1 . Prove that\frac{a}{2+b^{2}+c^{2}}+\frac{b}{2+c^{2}+a^{2}}+ \frac{c}{2+a^{2}+b^{2}} \geq \frac{3\sqrt{3}}{8}(Vietnam)
25.05.2017 15:50
Ji Chen wrote: Vasc wrote: The best inequality of this type is the following: If a,b,c are real numbers such that a + b + c = 3, then \frac 1{8 + 5(b^2 + c^2)} + \frac 1{8 + 5(c^2 + a^2)} + \frac 1{8 + 5(a^2 + b^2)}\le \frac 1{6}, with equality for a = b = c, and again for \frac a{13} = b = c or any permutation. \frac{1}{6}-\sum{\frac{1}{8+5b^2+5c^2}} =\frac{125}{27}\prod{\frac{(b-c)^2}{8+5b^2+5c^2}}+\frac{5}{54}\sum{\frac{(1-5a)^2(b-c)^2}{(8+5c^2+5a^2)(8+5a^2+5b^2)}}\geq0. http://www.artofproblemsolving.com/Forum/viewtopic.php?t=359320 szl,note jichen this post..
04.06.2019 03:41
khashi70 wrote: Suppose that a, b, c be three positive real numbers such that a+b+c=3 . Prove that : \frac{1}{2+a^{2}+b^{2}}+\frac{1}{2+b^{2}+c^{2}}+\frac{1}{2+c^{2}+a^{2}} \leq \frac{3}{4} Proof : By C-S and AM-GM, \sum \frac{1}{2+a^2+b^2}=\frac{3}{2}-\frac{1}{2}\sum \frac{a^2+b^2}{2+a^2+b^2}\leq \frac{3}{2}-\sum \frac{(\sum \sqrt{a^2+b^2})^2}{4\sum a^2+12}\leq \frac{3}{2}-\sum \frac{2\sum a^2+\sum (a+b)(a+c)}{4\sum a^2+12}=\frac{3}{2}-\sum \frac{3\sum a^2+3\sum ab}{4\sum a^2+12}\leq \frac{3}{2}-\sum \frac{3\sum a^2+9}{4\sum a^2+12}=\frac{3}{4}.
13.02.2020 09:23
Suppose that a, b, c be three positive real numbers such that a+b+c=3 . Prove that\frac{1}{a^2-b +4}+\frac{1}{b^2-c +4}+\frac{1}{c^2-a +4} \ge \frac{3}{4}p/6487921617 Suppose that a, b, c be three positive real numbers such that a+b+c=3 . Prove that\frac {1}{a^2+b^2+3} + \frac {1}{b^2+c^2+3} +\frac{1 } {c^2+a^2+3}\leq \frac {3}{5}Suppose that a, b, c be three positive real numbers such that abc(a+b+c)=3 . Prove that\frac{1}{a^2+b^2+1}+\frac{1}{b^2+c^2+1}+ \frac{1}{c^2+a^2+1}\leq 1
09.04.2020 08:26
sqing wrote: Suppose that a, b, c be three positive real numbers such that a+b+c=3 . Prove that\frac{1}{a^2-b +4}+\frac{1}{b^2-c +4}+\frac{1}{c^2-a +4} \ge \frac{3}{4} here
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26.04.2020 10:58
Let a,b,c>0 and ab+bc+ca=1. Prove that \frac{1}{1+a^2+b^2}+\frac{1}{1+b^2+c^2}+\frac{1}{1+c^2+a^2}\leq\frac{9}{5}Let a,b,c>0 and a^2+b^2+c^2=1. Prove that \frac{1}{1+a^2+b^2}+\frac{1}{1+b^2+c^2}+\frac{1}{1+c^2+a^2}\geq\frac{9}{5}Let a,b,c>0 and a+b+c=1. Prove that \frac{1}{1+a^2+b^2}+\frac{1}{1+b^2+c^2}+\frac{1}{1+c^2+a^2}\leq\frac{27}{11}Let a,b,c>0 and abc=1. Prove that \frac{1}{1+a^2+b^2}+\frac{1}{1+b^2+c^2}+\frac{1}{1+c^2+a^2}\leq1Let a,b,c>0 and a+b+c=2.Prove that \frac{1}{1+a^2+b^2}+\frac{1}{1+b^2+c^2}+\frac{1}{1+c^2+a^2}\geq\frac{4}{3}Suppose that a, b, c be three positive real numbers such that a+b+c=3 . Prove that \frac{a^2}{b^2-b+1}+\frac{b^2}{c^2-c+1}+\frac{c^2}{a^2-a+1}\geq 3
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14.06.2020 00:16
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14.06.2020 01:50
khashi70 wrote: Suppose that a, b, c be three positive real numbers such that a+b+c=3 . Prove that : \frac{1}{2+a^{2}+b^{2}}+\frac{1}{2+b^{2}+c^{2}}+\frac{1}{2+c^{2}+a^{2}} \leq \frac{3}{4} Because: \begin{align*} &571536\, \prod \left( {a}^{2}+{b}^{2}+2 \right) \cdot (\text{RHS}-\text{LHS}) \\&=3\sum\limits_{cyc} \, \left( 1524\,{a}^{2}+32\,ac+1780\,{b}^{2}+63\,{c}^{2} \right) \left( a+b-2\,c \right) ^{4}\\&\,\,\,+81\, \sum\limits_{cyc} \left( 1859\,{a}^{2}{b}^{2}+503\,ab {c}^{2}+3619\,a{c}^{3}+96\,{b}^{3}c+3331\,b{c}^{3} \right) \left( a-b \right) ^{2} \\& \geq 0\end{align*}
15.01.2021 16:49
khashi70 wrote: Suppose that a, b, c be three positive real numbers such that a+b+c=3 . Prove that : \frac{1}{2+a^{2}+b^{2}}+\frac{1}{2+b^{2}+c^{2}}+\frac{1}{2+c^{2}+a^{2}} \leq \frac{3}{4} Solution of Zhangyanzong: \sum_{cyc}\frac{a^2+b^2}{2+a^2+b^2}\geq \frac{(\sum_{cyc}\sqrt{a^2+b^2})^2}{2\sum_{cyc} a^2+6}=\frac{2\sum_{cyc} a^2+2\sum_{cyc}\sqrt{(a^2+b^2)(a^2+c^2)}}{2\sum_{cyc} a^2+6}\geq\frac{2\sum_{cyc} a^2+2\sum_{cyc}(a^2+bc)}{2\sum_{cyc} a^2+6} =\frac{3}{2} \sum_{cyc}\frac{1}{2+a^2+b^2}\leq \frac{3}{4}https://artofproblemsolving.com/community/c6h375652p19876954 Suppose that a, b, c be three positive real numbers such that a+b+c=3 . Prove that :
If a, b and c are positive then\sum\limits_{cyc} \frac{a^{2}}{b^{2}+c^{2}+bc}\geq 1
12.06.2022 09:24
Let a,b,c>0 and \frac{1}{a^2+b^2+2}+ \frac{1}{b^2+c^2+2}+\frac{1}{c^2+a^2+2}=1 . Prove that a+b+c\leq \frac{3}{\sqrt 2}Let a,b>0 and \frac{1}{a^2+b^2+2}+\frac{1}{a^2+2}+\frac{1}{b^2+2} =1 . Prove that a+b\leq \sqrt{\sqrt{17} -1} ab\leq \frac{\sqrt{17}-1}{4}here
10.10.2022 15:16
Let a,b,c>0 and \frac{1}{a^2+b^2+1}+ \frac{1}{b^2+c^2+1}+\frac{1}{c^2+a^2+1}\geq 1 . Prove that b^2+3 \ge 2b(a+c)ab + bc + ca \leq 3here
11.07.2023 04:18
Let a,b,c are nonnegative real numbers such that a + b + c = 3 . Prove that \frac 1{8 + 5(a^2 + b^2)} + \frac 1{8 + 5(b^2 + c^2)} + \frac 1{8 + 5(c^2 + a^2)}\geq \frac{642}{4697}with equality for (a,b,c)=\left(\frac 3{2},\frac 3{2},0\right). https://artofproblemsolving.com/community/c4h3108842p28127400
24.11.2024 16:59
Hope this is correct... We replace b^2+c^2=2x,a^2+c^2=2y,a^2+b^2=2z which yields 3xyz +\sum{xy}\overset{?}{\geq} \sum{x}+3with condition \sqrt{y+z-x}+\sqrt{x+z-y}+\sqrt{x+y-z}=3. This one is equavilent to (with different condition) \sum{\sqrt{x}}=3\overset{?}{\implies} 3\Pi{(y+z)}+2\sum{x^2}+6\sum{yz}\overset{?}{\geq} 8\sum{x}+24Let x+y+z=3u,xy+yz+zx=3v^2,xyz=w^3. 3(9uv^2-w^3)+2(9u^2-6v^2)+18v^2\overset{?}{\geq} 24u+24 \iff 9uv^2+6u^2+2v^2-8u-8\overset{?}{\geq} w^3We have u\geq 1 and u+v^2\geq 2. If u\geq 2, then 9uv^2+6u^2+2v^2-8u-8\geq w^3+12u+2v^2-8u-8\geq w^3+4u-8\geq w^3Suppose that u<2. 9uv^2+6u^2+2v^2-8u-8-w^3\geq 8uv^2+6u^2+2v^2-8u-8\overset{?}{\geq} 04uv^2+3u^2+v^2\overset{?}{\geq} 4u+4 \ \ \text{or} \ \ 4uv^2+3u^2+v^2\geq (4u+1)(2-u)\overset{?}{\geq} -3u^2+4u+4-4u^2-u+8u+2\overset{?}{\geq} -3u^2+4u+4\iff u^2-3u+2=(u-1)(u-2)\overset{?}{\leq} 0Which is true by our assumption as desired.\blacksquare Thanks below!
25.11.2024 01:07
bin_sherlo wrote: \sum{\frac{1}{b^2+c^2+1+1}}\overset{CS}{\leq} \sum{\frac{1+1+a+1}{(a+b+c+1)^2}}=\frac{\sum{a}+9}{16}=\frac{3}{4}As desired.\blacksquare I think the C-S is (b^2+c^2+1+1)(1+1+a^2+1) \geq (a+b+c+1)^2,And since a^2+b^2+c^2 \geq 3, this isn't strong enough.