By Vasc's RCF-Theorem it's enough to check only one case $b=c$ and $a=\frac{1}{b^2}$.

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2799770#p2799770 I can't write urls...

See here my proof.

Sorry if I'm being an idiot, but even though I now know the RCF, LCF and HCF theorems, can someone please explicitly explain how the RCF and HCF theorems can be applied to solve this problem? What I mean is explicitly stating why we are done if we can prove it for $b=c$. Thanks!

Sorry if I'm being an idiot , but what is the RCF, LCF, and HCF theorems?

Anyone who has an elementary solution without such theorems, as they aren't elementary or known. Otherwise it is bad as preparation on IMO where they don't may use it? * May countries as Croatia participate too?

SCP wrote: Anyone who has an elementary solution without such theorems, as they aren't elementary or known. Otherwise it is bad as preparation on IMO where they don't may use it? * May countries as Croatia participate too? Lemma. If $3ab\geq x\geq 0$, then \[\sqrt{x+a^{2}}+\sqrt{x+b^{2}}\geq 2\sqrt{x+ab}+\left( \sqrt{a}-\sqrt{b}\right) ^{2}.\] Proof. Since $\sqrt{\left( x+a^{2}\right) \left( x+b^{2}\right) }\geq \left(x+ab\right) $ and $2\sqrt{ab}\geq \sqrt{x+ab}$ hence \begin{eqnarray*} &&\left( \sqrt{x+a^{2}}+\sqrt{x+b^{2}}\right) ^{2}-\left( 2\sqrt{x+ab}+\left( \sqrt{a}-\sqrt{b}\right) ^{2}\right) ^{2} \\ &=&x+a^{2}+x+b^{2}+2\sqrt{\left( x+a^{2}\right) \left( x+b^{2}\right) } \\ &&-\left( 4\left( x+ab\right) +\left( \sqrt{a}-\sqrt{b}\right) ^{4}+2\sqrt{x+ab}\left( \sqrt{a}-\sqrt{b}\right) ^{2}\right) \\ &\geq &x+a^{2}+x+b^{2}+2\left( x+ab\right) \\ &&-\left( 4\left( x+ab\right) +\left( \sqrt{a}-\sqrt{b}\right) ^{4}+4\sqrt{ab}\left( \sqrt{a}-\sqrt{b}\right) ^{2}\right) \\&=&0.\end{eqnarray*} Now let $abc=1$ and $a\geq b\geq c>0$. We have $3ab\geq 3>\frac{9}{16}$. The inequality from the lemma \[ \sqrt{\frac{9}{16}+a^{2}}+\sqrt{\frac{9}{16}+b^{2}}\geq 2\sqrt{\frac{9}{16}+ab}+\left( \sqrt{a}-\sqrt{b}\right) ^{2} \] is equivalent to \[ f\left( a,b,c\right) \geq f\left( \sqrt{ab},\sqrt{ab},c\right) \] where \[f\left( a,b,c\right) =\sum_{cyc}\sqrt{9+16a^{2}}-3-4\sum_{cyc}a.\] The rest can be done as in MathUniverse's.

SCP wrote: Anyone who has an elementary solution without such theorems, as they aren't elementary or known. Otherwise it is bad as preparation on IMO where they don't may use it? * May countries as Croatia participate too? By the problem proposers this problem wasn't supposed to be solved by such methods and I believe you are mistaken on the point that such theorems are not allowed on the IMO they are but same as complex number solutions and analitic geometry there is a "rule" that you can get either 0 or 7 points using such tehniques . Croatia participates at MEMO for as long as it exists we Croatians like to believe that we are a part of Central Europe regardless of what the rest of the world believes we have even hosted the last year MEMO.

syk0526 wrote: Let $ a,b$ and $ c $ be positive real numbers with $ abc = 1 $. Prove that \[ \sqrt{ 9 + 16a^2}+\sqrt{ 9 + 16b^2}+\sqrt{ 9 + 16c^2} \ge 3 +4(a+b+c)\] Of course, there is another way to get along with this quite hard problem in a more or less non-olympic manner. Use the method of Lagrangian multipliers and thus deal with the function $\Phi(a,b,c,\lambda)= \sqrt{ 9 + 16a^2}+\sqrt{ 9 + 16b^2}+\sqrt{ 9 + 16c^2} -4(a+b+c)+\lambda(abc-1) $ Struggling through all necessary cases finally yields a sort of "standard" proof of the inequality. Indeed, it might be of some interest to study also inequalities of the more general nature \[ \sqrt{ p + qa^2}+\sqrt{ p + qb^2}+\sqrt{ p + qc^2} \ge ...\] with $p, q > 0$ to be specified (and of course $a, b, c > 0$ still satisfying $abc = 1$). Good luck wih this problem! auj

The following reverse inequality holds. Let $ a,b$ and $ c $ be positive real numbers with $ abc = 1 $. Prove that \[ \sqrt{ 144 + 25a^2}+\sqrt{ 144 + 25b^2}+\sqrt{ 144 + 25c^2} \le 24 +5(a+b+c).\]

Vasc wrote: The following reverse inequality holds. Let $ a,b$ and $ c $ be positive real numbers with $ abc = 1 $. Prove that \[ \sqrt{ 144 + 25a^2}+\sqrt{ 144 + 25b^2}+\sqrt{ 144 + 25c^2} \le 24 +5(a+b+c).\] Your LCF-Theorem helps here!

arqady wrote: Vasc wrote: The following reverse inequality holds. Let $ a,b$ and $ c $ be positive real numbers with $ abc = 1 $. Prove that \[ \sqrt{ 144 + 25a^2}+\sqrt{ 144 + 25b^2}+\sqrt{ 144 + 25c^2} \le 24 +5(a+b+c).\] Your LCF-Theorem helps here! I think he create this inequality with his theorem!!

Vasc wrote: The following reverse inequality holds. Let $ a,b$ and $ c $ be positive real numbers with $ abc = 1 $. Prove that \[ \sqrt{ 144 + 25a^2}+\sqrt{ 144 + 25b^2}+\sqrt{ 144 + 25c^2} \le 24 +5(a+b+c).\] Can it be proved like

bdtilove123 wrote: I think he create this inequality with his theorem!! It seems that you are right because the Can's idea $\sqrt{144+25b^2}+\sqrt{144+25c^2}\leq(b+c)\sqrt{144c+25}$ does not work here.

Vasc wrote: The following reverse inequality holds. Let $ a,b$ and $ c $ be positive real numbers with $ abc = 1 $. Prove that \[ \sqrt{ 144 + 25a^2}+\sqrt{ 144 + 25b^2}+\sqrt{ 144 + 25c^2} \le 24 +5(a+b+c).\][/qC uote] Could you please give an elementary solution without such theorems?... I am eager to know,but i can not finish it. sorry for my bad English.

This is nice. If $a,b,c$ are positive real numbers such that $abc=1$, then \[\sqrt{1-a+a^2}+\sqrt{1-b+b^2}+\sqrt{1-c+c^2}\ge a+b+c.\]

It seems that a nice proof for the last nice inequality is hard to find.

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Vasc wrote: The following reverse inequality holds. Let $ a,b$ and $ c $ be positive real numbers with $ abc = 1 $. Prove that \[ \sqrt{ 144 + 25a^2}+\sqrt{ 144 + 25b^2}+\sqrt{ 144 + 25c^2} \le 24 +5(a+b+c).\] Setting $ x=\sqrt{144+25a^2}-5a ; y=\sqrt{144+25b^2}-5b ; z=\sqrt{144+25c^2}-5c $ then $ 0 < x,y,z <12 $ and $ \dfrac{144-x^2}{10x} \cdot \dfrac{144-y^2}{10y} \cdot \dfrac{144-z^2}{10z}=abc=1 $ or $ (144-x^2)(144-y^2)(144-z^2)=10^3xyz $ And the inequality becomes $ x+y+z \leq 24 $ Contradict that $ x+y+z > 24 $ Then $ 144-x^2 < 6(x+y+z)-\dfrac{24x^2}{x+y+z}=\dfrac{6\left[(x+y+z)^2-4x^2 \right]}{x+y+z} $ Therefore \[ (144-x^2)(144-y^2)(144-z^2) \] \[ < \dfrac{6^3\left[(x+y+z)^2-4x^2 \right]\left[(x+y+z)^2-4y^2 \right]\left[(x+y+z)^2-4z^2 \right]}{(x+y+z)^3} \] \[ = \dfrac{6^3(y+z-x)(z+x-y)(x+y-z)(3x+y+z)(3y+z+x)(3z+x+y)}{(x+y+z)^3} \] \[ \leq \dfrac{6^3(y+z-x)(z+x-y)(x+y-z)}{(x+y+z)^3}\cdot \dfrac{5^3(x+y+z)^3}{27} \] \[ = 10^3(y+z-x)(z+x-y)(x+y-z) \leq 10^3xyz \] Which is false, so we have $ x+y+z \leq 24 $.It completed the proof.

Nice, quykhtn-qa1! However, there is at least a solution without HCF-Theorem or contradiction method.

Vasc wrote: Vasc wrote: The following reverse inequality holds. Let $ a,b$ and $ c $ be positive real numbers with $ abc = 1 $. Prove that \[ \sqrt{ 144 + 25a^2}+\sqrt{ 144 + 25b^2}+\sqrt{ 144 + 25c^2} \le 24 +5(a+b+c).\] Can somebody find a proof without using HCF-Theorem? I think there is at least a such proof. For this problem, I have two proofs: One using contradiction method as quykhtn has showed above, one will be present below: The original inequality is equivalent to \[\sum \left( \sqrt{25a^2+144}-5a\right) \le 24,\] or \[\sum \frac{12}{\sqrt{25a^2+144}+5a} \le 2.\] Now, we will show that for any positive real number $x,$ the following inequality holds: \[\frac{12}{\sqrt{25x^2+144}+5x} \le \frac{x^{\frac{10}{13}}+1}{x^{\frac{20}{13}}+x^{\frac{10}{13}}+1}.\quad (1)\] This inequality is equivalent to one of the following inequalities \[\sqrt{25x^2+144}+5x \ge\frac{12\left(x^{\frac{20}{13}}+x^{\frac{10}{13}}+1\right)}{x^{\frac{10}{13}}+1},\] \[\left(\sqrt{25x^2+144}-12\right)+5x \ge \frac{12x^{\frac{20}{13}}}{x^{\frac{10}{13}}+1},\] \[\frac{25x}{\sqrt{25x^2+144}+12}+5 \ge \frac{12x^{\frac{7}{13}}}{x^{\frac{10}{13}}+1}.\] Using the AM-GM inequality, we have \[\sqrt{25x^2+144}+12 \le \frac{1}{2}\left(5x+8+\frac{25x^2+144}{5x+8}\right) +12=\frac{25(x^2+4x+8)}{5x+8}.\] Therefore, it suffices to prove that \[\frac{x(5x+8)}{x^2+4x+8} +5\ge \frac{12x^{\frac{7}{13}}}{x^{\frac{10}{13}}+1}.\] This inequality is equivalent to \[\left(x^{\frac{10}{13}}+1\right)(10x^2+28x+40) \ge 12x^{\frac{7}{13}}(x^2+4x+8),\] or \[10x^{\frac{36}{13}}+10x^2+28x^{\frac{20}{13}}+28x+40x^{\frac{10}{13}}+40 \ge 12x^{\frac{33}{13}}+48x^{\frac{20}{13}}+96x^{\frac{7}{13}}.\quad (2)\] Using the AM-GM inequality, we have $10x^{\frac{36}{13}}+x^{\frac{23}{13}}+x \ge 12\sqrt[12]{x^{10\cdot \frac{36}{13}+\frac{23}{13}+1}}=12x^{\frac{33}{13}};$ $10x^{2}+23x^{\frac{23}{13}}+7x+8x^{\frac{10}{13}} \ge 48\sqrt[48]{x^{10\cdot 2+23\cdot \frac{23}{13}+7\cdot 1+8\cdot \frac{10}{13}}}=48x^{\frac{20}{13}};$ $4x^{\frac{23}{13}}+20x+32x^{\frac{10}{13}}+40 \ge 96\sqrt[96]{x^{4\cdot \frac{23}{13}+20\cdot 1+32\cdot \frac{10}{13}}}=96x^{\frac{7}{13}}.$ Adding these inequalities, we get $(2)$ and so, $(1)$ is proved. Now, from $(1),$ we see that it suffices to prove that \[\sum \frac{a^{\frac{10}{13}}+1}{a^{\frac{20}{13}}+a^{\frac{10}{13}}+1} \le 2,\] which is true according to the known result: If $x,\,y,\,z$ are real numbers satisfying $xyz=1,$ then \[\sum \frac{x+1}{x^2+x+1} \le 2.\]

Nice and interesting proof, ham-hap ! My proof is similar, but shorter. Since \[\sqrt{25a^2+144}-5a=\frac {144}{\sqrt{25a^2+144}+5a},\] the inequality is equivalent to \[\sum \frac 1{\sqrt{25a^2+144}+5a}\le \frac 1{6}.\] If the inequality \[\frac 1{\sqrt{25a^2+144}+5a}\le \frac 1{6\sqrt{5a^{18/13}+4}}\] holds for all $a>0$, then it suffices to show that \[\sum\frac 1{\sqrt{5a^{18/13}+4}}\le 1,\] which is a known inequality . Therefore, using the substitution $x=a^{1/13}$, $x>0$, we only need to show that \[\sqrt{25x^{26}+144}+5x^{13}\ge 6\sqrt{5x^{18}+4}.\] By squaring, the inequality becomes \[10x^{13}(\sqrt{25x^{26}+144}+5x^{13}-18x^5)\ge 0.\] This is true if \[25x^{26}+144\ge (18x^5-5x^{13})^2,\] which is equivalent to \[5x^{18}+4\ge 9x^{10}.\] By the AM-GM inequality, the conclusion follows.

ham-hap wrote: Vasc wrote: The following reverse inequality holds. Let $ a,b$ and $ c $ be positive real numbers with $ abc = 1 $. Prove that \[ \sqrt{ 144 + 25a^2}+\sqrt{ 144 + 25b^2}+\sqrt{ 144 + 25c^2} \le 24 +5(a+b+c).\] Can somebody find a proof without using HCF-Theorem? I think there is at least a such proof. For this problem, I have two proofs: One using contradiction method as quykhtn has showed above, one will be present below: The original inequality is equivalent to \[\sum \left( \sqrt{25a^2+144}-5a\right) \le 24,\] or \[\sum \frac{12}{\sqrt{25a^2+144}+5a} \le 2......\] My second proof is similar with ham-hap ' proof. The original inequality is equivalent to \[\sum \left(\sqrt{25a^2+144}-5a\right) \leq 24,\] or \[\sum \frac{6}{\sqrt{25a^2+144}+5a} \le 1.\] Since $ \sqrt{25a^2+144}<\sqrt{25a^2+120a+144}=5a+12 $,we have \[ \sqrt{25a^2+144}-\frac{25a+144}{13} =\frac{13\sqrt{25a^2+144}-(25a+144)}{13} \] \[ =\dfrac{3600(a-1)^2}{13\left[13\sqrt{25a^2+144}+(25a+144)\right]}\geq \frac{3600(a-1)^2}{13\left[13(5a+12)+(25a+144) \right]} \] \[=\frac{120(a-1)^2}{13(3a+10)} \] So, $ \sqrt{25a^2+144}+5a \geq \dfrac{25a+144}{13}+\dfrac{120(a-1)^2}{13(3a+10)}+5a=\dfrac{6(5a^2+14a+20)}{3a+10} $ Therefore, it suffices to prove that \[\sum \frac{3a+10}{5a^2+14a+20} \leq 1.\] or \[ \sum \left[\dfrac{1}{2}-\dfrac{3a+10}{5a^2+14a+20}\right] \geq \dfrac{1}{2}; \] or \[ \sum \dfrac{5a^2+8a}{5a^2+14a+20} \geq 1 \] Now, we will prove that for any positive real number $x,$ the following inequality holds \[ \dfrac{5x^2+8x}{5x^2+14x+20} \geq \dfrac{x^{\dfrac{20}{13}}}{x^{\dfrac{20}{13}}+x^{\dfrac{10}{13}}+1} \quad (1) \] or \[ (5x+8)\left(x^{\dfrac{20}{13}}+x^{\dfrac{10}{13}}+1\right) \geq x^{\dfrac{7}{13}}(5x^2+14x+20),\] or \[ 5x^{\frac{23}{13}}+5x+8x^{\frac{10}{13}}+8 \geq 6x^{\frac{20}{13}}+20x^{\frac{7}{13}}.\] Using the AM-GM inequality, we see that \[ 4x^{\frac{23}{13}}+x^{\frac{15}{13}}+x \ge 6\sqrt[6]{x^{4\cdot \frac{23}{13}+\frac{15}{13}+1}}=6x^{\frac{20}{13}}; \] \[ 8x^{\frac{10}{13}}+8+4x^{\frac{15}{13}} \ge 20\sqrt[20]{x^{8\cdot \frac{10}{13}+4\cdot \frac{15}{13}}}=20x^{\frac{7}{13}}.\] and \[ x^{\frac{23}{13}}+4x \ge 5\sqrt[5]{x^{\frac{23}{13}+4}}=5x^{\frac{15}{13}} \] Therefore, the last inequality is true and (1) is proved. Now, using (1), it remains to prove that \[ \sum\frac{a^{\frac{20}{13}}}{a^{\frac{20}{13}}+a^{\frac{10}{13}}+1} \ge 1,\] which is true according to the known result: If $x,\,y,\,z$ are real numbers satisfying $xyz=1,$ then \[ \dfrac{x^2}{x^2+x+1}+\dfrac{y^2}{y^2+y+1}+\dfrac{z^2}{z^2+z+1} \ge 1.\]

Vasc wrote: This is nice. If $a,b,c$ are positive real numbers such that $abc=1$, then \[\sqrt{1-a+a^2}+\sqrt{1-b+b^2}+\sqrt{1-c+c^2}\ge a+b+c.\] Here is a simple solution. We will prove that for any positive real number $x,$ the following inequality holds \[ \sqrt{x^2-x+1}-x \geq \dfrac{1}{2}\left(\dfrac{3}{x^2+x+1}-1 \right) .\quad (1) \] This inequality is equivalent to \[ 2\sqrt{x^2-x+1} \geq \dfrac{2x^3+x^2+x+2}{x^2+x+1} \] Since $ 4(x^2-x+1)(x^2+x+1)^2-(2x^3+x^2+x+2)^2=3x^4-6x^3+3x^2=3x^2(x-1)^2 \geq 0 $ Therefore, the last inequality is true and (1) is proved. Now, using (1), it remains to prove that \[ \dfrac{1}{a^2+a+1}+\dfrac{1}{b^2+b+1}+\dfrac{1}{c^2+c+1} \geq 1 \] Which is the known result.

Nice proof, quykhtn-qa1! Simply, the following inequalities are equivalent. \[ \sqrt{x^2-x+1}-x \geq \dfrac{1}{2}\left(\dfrac{3}{x^2+x+1}-1 \right) , \] \[\frac{1-x}{\sqrt{x^2-x+1}+x}\ge \frac{(1-x)(2+x)}{2(x^2+x+1)},\] \[(x-1)[ (x+2)\sqrt{x^2-x+1}-x^2-2]\ge 0,\] \[\frac{3x^2(x-1)^2}{(x+2)\sqrt{x^2-x+1}+x^2+2}\ge 0.\]

syk0526 wrote: Let $ a,b$ and $ c $ be positive real numbers with $ abc = 1 $. Prove that \[ \sqrt{ 9 + 16a^2}+\sqrt{ 9 + 16b^2}+\sqrt{ 9 + 16c^2} \ge 3 +4(a+b+c)\] Kimpul wrote: Lemma. If $3ab\geq x\geq 0$, then \[\sqrt{x+a^{2}}+\sqrt{x+b^{2}}\geq 2\sqrt{x+ab}+\left( \sqrt{a}-\sqrt{b}\right) ^{2}.\] Proof. Since $\sqrt{\left( x+a^{2}\right) \left( x+b^{2}\right) }\geq \left(x+ab\right) $ and $2\sqrt{ab}\geq \sqrt{x+ab}$ hence \begin{eqnarray*} &&\left( \sqrt{x+a^{2}}+\sqrt{x+b^{2}}\right) ^{2}-\left( 2\sqrt{x+ab}+\left( \sqrt{a}-\sqrt{b}\right) ^{2}\right) ^{2} \\ &=&x+a^{2}+x+b^{2}+2\sqrt{\left( x+a^{2}\right) \left( x+b^{2}\right) } \\ &&-\left( 4\left( x+ab\right) +\left( \sqrt{a}-\sqrt{b}\right) ^{4}+2\sqrt{x+ab}\left( \sqrt{a}-\sqrt{b}\right) ^{2}\right) \\ &\geq &x+a^{2}+x+b^{2}+2\left( x+ab\right) \\ &&-\left( 4\left( x+ab\right) +\left( \sqrt{a}-\sqrt{b}\right) ^{4}+4\sqrt{ab}\left( \sqrt{a}-\sqrt{b}\right) ^{2}\right) \\&=&0.\end{eqnarray*} Now let $abc=1$ and $a\geq b\geq c>0$. We have $3ab\geq 3>\frac{9}{16}$. The inequality from the lemma \[ \sqrt{\frac{9}{16}+a^{2}}+\sqrt{\frac{9}{16}+b^{2}}\geq 2\sqrt{\frac{9}{16}+ab}+\left( \sqrt{a}-\sqrt{b}\right) ^{2} \] is equivalent to \[ f\left( a,b,c\right) \geq f\left( \sqrt{ab},\sqrt{ab},c\right) \] where \[f\left( a,b,c\right) =\sum_{cyc}\sqrt{9+16a^{2}}-3-4\sum_{cyc}a.\] \[ \left( 2\sqrt{x+ab}+\left( \sqrt{a}-\sqrt{b}\right) ^{2}\right) ^{2}=\left( 4\left( x+ab\right) +\left( \sqrt{a}-\sqrt{b}\right) ^{4}+2\sqrt{x+ab}\left( \sqrt{a}-\sqrt{b}\right) ^{2}\right)\] This step is wrong ! It must be \[ \left( 2\sqrt{x+ab}+\left( \sqrt{a}-\sqrt{b}\right) ^{2}\right) ^{2}=\left( 4\left( x+ab\right) +\left( \sqrt{a}-\sqrt{b}\right) ^{4}+4\sqrt{x+ab}\left( \sqrt{a}-\sqrt{b}\right) ^{2}\right)\] So your proof is wrong.Anyone can post a full proof by Mixing variables or elementary solution by classical inequality?

Vasc wrote: The following reverse inequality holds. Let $ a,b$ and $ c $ be positive real numbers with $ abc = 1 $. Prove that \[ \sqrt{ 144 + 25a^2}+\sqrt{ 144 + 25b^2}+\sqrt{ 144 + 25c^2} \le 24 +5(a+b+c).\] Let $ a,b$ and $ c $ be positive real numbers with $ abc = 1 $. Prove that \[ \sqrt{ 145 + 24a^2}+\sqrt{ 145 + 24b^2}+\sqrt{ 145 + 24c^2} \le 24 +5(a+b+c).\]\[ \sqrt{ 146 + 23a^2}+\sqrt{ 146 + 23b^2}+\sqrt{ 146 + 23c^2} \le 24 +5(a+b+c).\]

Vasc wrote: This is nice. If $a,b,c$ are positive real numbers such that $abc=1$, then \[\sqrt{1-a+a^2}+\sqrt{1-b+b^2}+\sqrt{1-c+c^2}\ge a+b+c.\]

syk0526 wrote: Let $ a,b$ and $ c $ be positive real numbers with $ abc = 1 $. Prove that \[ \sqrt{ 9 + 16a^2}+\sqrt{ 9 + 16b^2}+\sqrt{ 9 + 16c^2} \ge 3 +4(a+b+c)\]

Vasc wrote: This is nice. If $a,b,c$ are positive real numbers such that $abc=1$, then \[\sqrt{1-a+a^2}+\sqrt{1-b+b^2}+\sqrt{1-c+c^2}\ge a+b+c.\] (Vasile Cîrtoaje, 2012)

Sqing, see the post #45

mudok wrote: Sqing, see the post #45 Sorry. Thanks.

http://epsilon.ro/wp-content/uploads/2012/09/2013_Matematica_Internationala_Concursul-MEMO2012_Solutii.pdf