Maximum is $\frac{1}{16}$

Easy to see that it suffices to consider $x=a+b>0$. We want to find the least value of the inverse. By C-S we have $$\frac{(4a^2+1+2)(1+4b^2+2)}{a+b}\ge\frac{4(a+b+1)^2}{a+b}=4\left(x+2+\frac{1}{x}\right)\ge 16$$

Taking the partial derivative of $E$ with respect to $a$ and $b$ and setting them equal to 0 gives $\frac{4a^2+3-8a(a+b)}{stuff}=0$, $\frac{4b^2+3-8a(a+b)}{stuff}=0$, where $stuff$ is always positive so the derivatives are always defined. Simplifying, we have $4a^2+8ab=4b^2+8ab=3$. Thus $a=b$ or $a=-b$. In the first case, $12a^2=3$ so $a=b=\frac{1}{2}, a=-\frac{1}{2}$. Testing those yields $E(a,b)=\frac{1}{16}$ as the max. In the second case, $E(a,b)=0$. Thus the maximum is $\frac{1}{16}$.

Steve12345 wrote: Find the maximum value of: $E(a,b)=\frac{a+b}{(4a^2+3)(4b^2+3)}$ For $a,b$ real numbers. Solution of Yechaojie：

Attachments:

Nice Inequalitie

Mathsy123 wrote: Nice Inequalitie Thank you https://ssmr.ro/files/onm2019/subiecte/ONM_2019_Juniori1.pdf and all Romanian ONM 2019 is here: https://ssmr.ro/onm2019

Good:

By AM-GM, $a+b=(a+b)\cdot 1\leq \left(\frac{a+b+1}{2}\right)^2$ By C-S, $(4a^2+3)(4b^2+3)=(4a^2+1+2)(1+4b^2+2)\geq (2a+2b+2)^2$ $$\frac{a+b}{(4a^2+3)(4b^2+3)}\leq\frac{(\frac{a+b+1}{2})^2}{(2a+2b+2)^2}=\frac{1}{16}.$$Equality holds when $a=b=\frac{1}{2}.$

Let $a,b>0 $ and $3(a^2+b^2-1)=4(a+b).$ Find the maximum value of $\frac{16}{a}+\frac{1}{b}.$

Let $a,b,c$ be real numbers such that $a^2+b^2+c^2=2.$ Prove that $$a+b+c\leq abc+2.$$Polish 1991

Let $a,b,c,d$ be real numbers such that $|a|,|b|,|c|,|d|>1 $ and $abc+bcd+cda+dab+a+b+c+d=0.$ Prove that $$\frac{1}{a-1}+\frac{1}{b-1}+\frac{1}{c-1}+\frac{1}{d-1}>0.$$

Let $a,b,c,d$ be positive real numbers such that $a^2+b^2+c^2+d^2=4.$ Prove that $$\frac{a+b+c+d}{2}\geq 1+\sqrt{abcd}.$$

Attachments:

Let $a,b,c$ be real numbers such that $ab+bc+ca=0.$ Prove that $$2(a^2+b^2+c^2)(a^2b^2+b^2c^2+c^2a^2)\geq 27a^2b^2c^2.$$

sqing wrote: Let $a,b>0 $ and $3(a^2+b^2-1)=4(a+b).$ Find the maximum value of $\frac{16}{a}+\frac{1}{b}.$ Let $a,b>0 $ and $3(a^2+b^2-1)=4(a+b).$ Prove that$$\frac{16}{a}+\frac{1}{b}\geq 9.$$Equality holds when $a=2,b=1.$

sqing wrote: Let $a,b,c$ be real numbers such that $ab+bc+ca=0.$ Prove that $$2(a^2+b^2+c^2)(a^2b^2+b^2c^2+c^2a^2)\geq 27a^2b^2c^2.$$ Do you have a solution please, because I am stuck and I need to see a way to solve it?

sqing wrote: Let $a,b,c$ be real numbers such that $ab+bc+ca=0.$ Prove that $$2(a^2+b^2+c^2)(a^2b^2+b^2c^2+c^2a^2)\geq 27a^2b^2c^2.$$ We have \[(a-b)^2(b-c)^2(c-a)^2 = \left[2(a^2+b^2+c^2)(a^2b^2+b^2c^2+c^2a^2) - 27a^2b^2c^2\right]-(a+b+c)(ab+bc+ca)\left[(a+b+c)(ab+bc+ca)-10abc\right].\]Let $ab+bc+ca=0$ then \[0 \leqslant (a-b)^2(b-c)^2(c-a)^2 = 2(a^2+b^2+c^2)(a^2b^2+b^2c^2+c^2a^2) - 27a^2b^2c^2.\]Therefore \[2(a^2+b^2+c^2)(a^2b^2+b^2c^2+c^2a^2) \geqslant 27a^2b^2c^2.\]Done.

Thank you, Song and Nguyen. #14 is mine and Valmir's (see Kosovo 2019 xii-th grade, it's an easy refinement) #15 is mine too. It was designed as P_1 in TST 5 (the easiest), while #14 was P_2 in TST 4 (Romania for JBMO 2019). See the references: https://pregatirematematicaolimpiadejuniori.files.wordpress.com/2019/05/2019_baraj_juniori-4_solutii.pdf https://pregatirematematicaolimpiadejuniori.files.wordpress.com/2019/05/2019_baraj_juniori-5_solutii.pdf

Nguyenhuyen_AG wrote: sqing wrote: Let $a,b,c$ be real numbers such that $ab+bc+ca=0.$ Prove that $$2(a^2+b^2+c^2)(a^2b^2+b^2c^2+c^2a^2)\geq 27a^2b^2c^2.$$ We have \[(a-b)^2(b-c)^2(c-a)^2 = \left[2(a^2+b^2+c^2)(a^2b^2+b^2c^2+c^2a^2) - 27a^2b^2c^2\right]-(a+b+c)(ab+bc+ca)\left[(a+b+c)(ab+bc+ca)-10abc\right].\]Let $ab+bc+ca=0$ then \[0 \leqslant (a-b)^2(b-c)^2(c-a)^2 = 2(a^2+b^2+c^2)(a^2b^2+b^2c^2+c^2a^2) - 27a^2b^2c^2.\]Therefore \[2(a^2+b^2+c^2)(a^2b^2+b^2c^2+c^2a^2) \geqslant 27a^2b^2c^2.\]Done. Thank you! Indeed, the factorisation was too difficult for me to find it.

sqing wrote: Let $a,b,c$ be real numbers such that $ab+bc+ca=0.$ Prove that $$2(a^2+b^2+c^2)(a^2b^2+b^2c^2+c^2a^2)\geq 27a^2b^2c^2.$$ This is the solution I provided. Its purpose was the small degree of difficulty.

Attachments:

sqing wrote: Let $a,b,c,d$ be positive real numbers such that $a^2+b^2+c^2+d^2=4.$ Prove that $$\frac{a+b+c+d}{2}\geq 1+\sqrt{abcd}.$$ The solution by the authors. Again, we didn't consider MacLaurin inequality or AM-GM for more than 2 variables.

Attachments:

sqing wrote: Let $a,b,c$ be real numbers such that $ab+bc+ca=0.$ Prove that $$2(a^2+b^2+c^2)(a^2b^2+b^2c^2+c^2a^2)\geq 27a^2b^2c^2.$$#15 Let $a,b,c\in \mathbb{R}$ such that $ab+bc+ca=0$. Prove that $$\dfrac{a^2}{bc}+\dfrac{b^2}{ca}+\dfrac{c^2}{ab} \leq \dfrac{-15}{4}$$Given $a+b+c=0$ prove: $$27a^2b^2c^2+4(ab+bc+ca)^3\leq0$$Let $a,b,c$ is nonzero real number that $a+b+c=0.$prove: $$\frac{a^2b^2c^2}{(a^2+ab+b^2)(b^2+bc+c^2)(c^2+ca+a^2)}\leq\frac{4}{27}.$$

sqing wrote: Let $a,b,c$ be real numbers such that $ab+bc+ca=0.$ Prove that $$2(a^2+b^2+c^2)(a^2b^2+b^2c^2+c^2a^2)\geq 27a^2b^2c^2.$$ On expanding we get, $$6a^2b^2c^2+2\sum_{sym}(a^4b^2)\ge 27a^2b^2c^2$$or $$2\sum_{sym}(a^4b^2)\ge 22a^2b^2c^2$$. since $ab+bc+ca=0,$ the inequality is equivalent to, $$(ab+bc+ca)^3+2\sum_{sym}(a^4b^2)\ge 22a^2b^2c^2$$or $$a^3b^3+b^3c^3+c^3a^3+\sum_{sym}(b^3a^2c)+3a^2b^2c^2+2\sum_{sym}(a^4b^2)\ge 22a^2b^2c^2.$$Since, $(4,2,0)$ majorizes $(2,2,2)$ therefore, $$[4,2,0]\geq[2,2,2]$$or $$\sum_{sym}(a^4b^2)\geq 6\cdot a^2b^2c^2.$$Similarly Since, $(3,3,0)$ majorizes $(2,2,2)$ therefore, $$[3,3,0]\geq[2,2,2]$$or $$\sum_{sym}(a^3b^3)\geq 3\cdot a^2b^2c^2.$$and, Since, $(3,2,1)$ majorizes $(2,2,2)$ therefore, $$[3,2,1]\geq[2,2,2]$$or $$\sum_{sym}(b^3a^2c^1)\geq 6\cdot a^2b^2c^2.$$Thus, from above we get. $$2\sum_{sym}(a^4b^2)+\sum_{sym}(a^3b^3)+\sum_{sym}(b^3a^2c^1) \geq 21a^2b^2c^2.$$or $$a^3b^3+b^3c^3+c^3a^3+ b^3a^2c+b^3c^2a+c^3a^2b+c^3b^2a+a^3c^2b+a^3b^2c+3a^2b^2c^2+2(a^4b^2+b^4a^2+b^4c^2+c^4b^2+c^4a^2+a^4c^2)\ge 24a^2b^2c^2.$$Thus, we need to prove that $$a^3b^3+b^3c^3+c^3a^3+\sum_{sym}(b^3a^2c)+3a^2b^2c^2+2\sum_{sym}(a^4b^2)\ge 24a^2b^2c^2\ge 22a^2b^2c^2$$or $$24a^2b^2c^2\ge 22a^2b^2c^2$$or $$2a^2b^2c^2 \ge 0$$which is obvious. this ends our proof.

sqing wrote: Let $a,b,c$ is nonzero real number that $a+b+c=0.$prove: $$\frac{a^2b^2c^2}{(a^2+ab+b^2)(b^2+bc+c^2)(c^2+ca+a^2)}\leq\frac{4}{27}.$$ Since. $$a+b+c=0$$, by AM-GM we have, $$\frac{a+b+c}{3}=0\geq \sqrt[3]{abc}$$or $$abc\leq 0.$$Thus, $$\frac{a^2b^2c^2}{(a^2+ab+b^2)(b^2+bc+c^2)(c^2+ca+a^2)}\le \frac{0}{(a^2+ab+b^2)(b^2+bc+c^2)(c^2+ca+a^2)}=0.$$so it suffices to prove $$0\leq \frac{4}{27}$$which is obvious. Thus, this ends our proof $\blacksquare $

sqing wrote: Given $a+b+c=0$ prove: $$27a^2b^2c^2+4(ab+bc+ca)^3\leq0$$ By AM-GM, we have, $$27a^2b^2c^2\leq 27(\frac{a+b+c}{3})^6=0$$and $$4(ab+bc+ca)^3\leq4\cdot \frac{(a+b+c)^6}{27}=0.$$Thus, $$27a^2b^2c^2+4(ab+bc+ca)^3\leq0$$which ends our proof. $\blacksquare$

Let a,b,c be reals such that $a^2+b^2+c^2=2. $ Prove that $$a+b+c-2abc\leq\frac{2\sqrt{9+\sqrt{6}}}{3}$$$$a+b+c-\frac{1}{2}abc\leq\frac{8\sqrt{6}}{9}$$ sqing wrote: Let $a,b,c$ be real numbers such that $a^2+b^2+c^2=2.$ Prove that $$a+b+c\leq abc+2.$$Polish 1991 Let a,b,c be reals such that $a^2+b^2+c^2=3. $ Prove that $$a+\frac{1}{2}b+c-abc\leq\frac{ 29}{6\sqrt{3}}$$https://artofproblemsolving.com/community/c6h3070906p27724088

sqing wrote: sqing wrote: Let $a,b>0 $ and $3(a^2+b^2-1)=4(a+b).$ Find the maximum value of $\frac{16}{a}+\frac{1}{b}.$ Let $a,b>0 $ and $3(a^2+b^2-1)=4(a+b).$ Prove that$$\frac{16}{a}+\frac{1}{b}\geq 9.$$Equality holds when $a=2,b=1.$ Bump

sqing wrote: Let $a,b,c,d$ be positive real numbers such that $a^2+b^2+c^2+d^2=4.$ Prove that $$\frac{a+b+c+d}{2}\geq 1+\sqrt{abcd}.$$ so let us assume that $a+b=m$, $c+d=n$, $ab=x$ and $cd=y$. Now, $$\frac{a+b+c+d}{2}\geq 1+\sqrt{abcd}$$is eequivalent to $$\frac{m+n}{2}\geq 1+\sqrt{xy}.$$Let us define $$f(x,y)=xy-(\frac{m+ n-2}{2}).$$Now, $$\frac{\partial^2 f(x,y)}{\partial^2 x}=0$$and $$\frac{\partial^2 f(x,y)}{\partial^2 y}=0$$. Thus $f(x,y)$ is convex in variables $x$ and $y$. since, $x=ab$ and $y=bc$, therefore $0\leq x\leq \frac{m^2}{4}$ and $0\leq y\leq \frac{n^2}{4}.$ According to the proposition that an n-variable convex function, when considered as a single variable function, it attains the maximum values at its boundaries. so, $$\max f(x,y)=\max f(\alpha,\beta)$$where $\alpha \in \{0, \frac{m^2}{4}\}$ and $\beta \in \{0, \frac{n^2}{4}\}.$ Now we need to consider the following two cases. Case - 1 : when $x=\frac{m^2}{4}$ and $y=\frac{n^2}{4}$, we get, $a=b$ and $c=d$. Thus, we need to prove the following inequality, $$a+c=1+ac$$when $a^2+c^2=2$. on squaring both sides we get, $$a^2+c^2=1+a^2c^2$$or $$(a^2-1)^2\geq 0$$which is obvious. Case - 2 : when $x=0$ and $y=0$ or $abcd=0$. This forces one of the variables to be equal to zero. Let us assume that $d=0$. So we need to prove the following, $$a+b+c\ge 2$$when $a^2+b^2+c^2=4$. When we square both the sides we get, $$(a+b+c)^2=a^2+b^2+c^2+2\sum(ab)\ge 4=a^2+b^2+c^2$$or $$2\sum(ab)\ge 0$$, which is obvious. This ends our proof. $\blacksquare$

sqing wrote: sqing wrote: Let $a,b>0 $ and $3(a^2+b^2-1)=4(a+b).$ Find the maximum value of $\frac{16}{a}+\frac{1}{b}.$ Let $a,b>0 $ and $3(a^2+b^2-1)=4(a+b).$ Prove that$$\frac{16}{a}+\frac{1}{b}\geq 9.$$Equality holds when $a=2,b=1.$ We have $$4(a+b)=3(a^2+b^2-1)\Leftrightarrow 2b+8=(a-2)^2+(b-1)^2+2(a^2+b^2).$$ Then, $2b+8\geq 2(a^2+b^2)\Rightarrow b+4\geq a^2+b^2\Rightarrow b+4\geq\frac{4}{3}(a+b)+1\Rightarrow 3b+12\geq 4a+4b+3\Rightarrow 9\geq 4a+b$. Finally, by Cauchy-Schwartz, $$((\frac{8}{2\sqrt{a}})^2+(\frac{1}{\sqrt{b}})^2)((2\sqrt{a})^2+(\sqrt{b})^2)\geq (8+1)^2\Rightarrow (\frac{16}{a}+\frac{1}{b})(4a+b)\geq 9^2\Rightarrow \frac{16}{a}+\frac{1}{b}\geq \frac{9^2}{4a+b}\geq \frac{9^2}{9}=9$$as desired.