Clearly, it follows from Cauchy-Schwarz inequality.

cezar lupu wrote: Clearly, it follows from Cauchy-Schwarz inequality. Again you start the same bad behavior. If you don't post a full solution don't post anything. maybe you are a genius and you understand all things with a first sight and with only a hint. If yes congratulations you are the best.

silouan wrote: cezar lupu wrote: Clearly, it follows from Cauchy-Schwarz inequality. Again you start the same bad behaviour. If you don't post a full solution don't post anything. Agreed... however a little frustration a day seems to be healthy. (solution split off into new post for resource project)

By Cauchy-Schwarz we have $\left(\sum_{cyc}\frac{a}{b+2c+3d}\right)\left(\sum_{cyc}a(b+2c+3d)\right) \ge \left(\sum_{cyc}a\right)^2$ So our LHS is greater than $\frac{(a+b+c+d)^2}{4(ab+ac+ad+bc+bd+cd)}\ge\frac{2}{3}$, where the latter follows from 6 times AM-GM on the numerator. Alternatively one can apply Jensen on the convex function $f(x)=\frac1x$ to obtain the same results.

Peter wrote: By Cauchy-Schwarz we have $ \left(\sum_{cyc}\frac{a}{b+2c+3d}\right)\left(\sum_{cyc}a(b+2c+3d)\right)\ge\left(\sum_{cyc}a\right)^{2}$ So our LHS is greater than $ \frac{(a+b+c+d)^{2}}{4(ab+ac+ad+bc+bd+cd)}\ge\frac{2}{3}$, where the latter follows from 6 times AM-GM on the numerator. Alternatively one can apply Jensen on the convex function $ f(x) =\frac{1}{x}$ to obtain the same results. $ (a+b+c+d)^{2}= 2(ab+bc+cd+da+ac+bd)+a^{2}+b^{2}+c^{2}+d^{2}\geq\frac{8}{3}(ab+bc+cd+da+ac+bd)$ So,we have $ \frac{(a+b+c+d)^{2}}{4(ab+ac+ad+bc+bd+cd)}\ge\frac{2}{3}$

Peter wrote: By Cauchy-Schwarz we have $\left(\sum_{cyc}\frac{a}{b+2c+3d}\right)\left(\sum_{cyc}a(b+2c+3d)\right) \ge \left(\sum_{cyc}a\right)^2$ So our LHS is greater than $\frac{(a+b+c+d)^2}{4(ab+ac+ad+bc+bd+cd)}\ge\frac{2}{3}$, where the latter follows from 6 times AM-GM on the numerator. Alternatively one can apply Jensen on the convex function $f(x)=\frac1x$ to obtain the same results. well my solution with Cauchy Schwarz was same. But i am a bit uncomfortable with Jensen's. So, could you elaborate that solution too.. ??

Fermat -Euler wrote: Prove that \[ \frac{a}{b+2c+3d} +\frac{b}{c+2d+3a} +\frac{c}{d+2a+3b}+ \frac{d}{a+2b+3c} \geq \frac{2}{3} \] for all positive real numbers $a,b,c,d$. Using Titu's lemma we get, \[ \frac{a^2}{a(b+2c+3d)} +\frac{b^2}{b(c+2d+3a)} +\frac{c^2}{c(d+2a+3b)}+ \frac{d^2}{d(a+2b+3c)} \geq \frac{(a+b+c+d)^2}{4(ab+ac+ad+bc+bd+cd)}.\] From here it is sufficient to prove that, \[\frac{(a+b+c+d)^2}{4(ab+ac+ad+bc+bd+cd)}\ge \frac{2}{3}.\] Cross multiplying we get, \[3(a+b+c+d)^2\ge 8(ab+ac+ad+bc+bd+cd).\] Expanding $3(a+b+c+d)^2$, rearranging terms, and adding $a^2+b^2+c^2+d^2$ to both sides we get, \[4(a^2+b^2+c^2+d^2) \ge (a+b+c+d)^2.\] Since this is true by Cauchy-Schwarz, we are done. $\blacksquare$

Quite easy ISL, once you know Titu Lemma. Multiply numerator and denominator of each fraction by its numerator. $LHS\geq\frac{(a+b+c+d)^2}{4(ab+bc+cd+da+ac+bd)}$ We need to prove $3(a+b+c+d)^2\geq 8(ab+ac+ad+bc+bd+cd)$. Expand RHS, and we get something obvious by C-S

Fermat -Euler wrote: Prove that \[ \frac{a}{b+2c+3d} +\frac{b}{c+2d+3a} +\frac{c}{d+2a+3b}+ \frac{d}{a+2b+3c} \geq \frac{2}{3} \]for all positive real numbers $a,b,c,d$. Prove that $$\frac{a}{4a+5b+c+5d}+ \frac{b}{4b+5c+d+5a}+\frac{c}{4c+5d+a+5b}+\frac{d}{4d+5a+b+5c}\leq \frac{2}{5}$$for all positive real numbers $a,b,c,d.$

sqing wrote: Fermat -Euler wrote: Prove that \[ \frac{a}{b+2c+3d} +\frac{b}{c+2d+3a} +\frac{c}{d+2a+3b}+ \frac{d}{a+2b+3c} \geq \frac{2}{3} \]for all positive real numbers $a,b,c,d$. Prove that $$\frac{a}{4a+5b+c+5d}+ \frac{b}{4b+5c+d+5a}+\frac{c}{4c+5d+a+5b}+\frac{d}{4d+5a+b+5c}\leq \frac{2}{5}$$for all positive real numbers $a,b,c,d.$ shouldnt be the sign on the opposite side?

Sir what is Jensen pl tell

Vidush wrote: Sir what is Jensen pl tell Jensen Inequality https://en.m.wikipedia.org/wiki/Jensen%27s_inequality

Vidush wrote: Sir what is Jensen pl tell If a function is convex in an interval $A\subseteq \mathbb{R}$($f''(x)>0$ for every $x\in A$) then for every $a_i\in A$ holds that $\frac{f(a_1)+f(a_2)+...+f(a_n)}{n}\geq f(\frac{a_1+a_2+...+a_n}{n})$. If a function is concave in an interval $A \subseteq \mathbb{R}$($f''(x)<0$ for every $x\in A$) then for every $a_i\in A$ holds that$\frac{f(a_1)+f(a_2)+...+f(a_n)}{n}\leq f(\frac{a_1+a_2+...+a_n}{n})$

Could someone post a solution using Jensen's with $f(x)=\frac{1}{x}$?

isaacmeng wrote: Could someone post a solution using Jensen's with $f(x)=\frac{1}{x}$? Using the weighted form of Jensen, $$\sum_{\text{cyc}} (\frac{a}{a+b+c+d} \cdot \frac{1}{b+2c+3d}) \geq \frac{1}{\frac{1}{a+b+c+d} \cdot (\sum_{\text{cyc}} a(b+2c+3d))} = \frac{a+b+c+d}{4(ab+ac+ad+bc+bd+cd)}$$$$\Rightarrow \sum_{\text{cyc}} \frac{a}{b+2c+3d} \geq \frac{(a+b+c+d)^2}{4(ab+ac+ad+bc+bd+cd)}$$and then proceed as in the above solutions.

Cezar Lupu wrote: Clearly, it follows from Cauchy-Schwarz inequality. whats that?(too lazy to spell it out)

jacoporizzo wrote: isaacmeng wrote: Could someone post a solution using Jensen's with $f(x)=\frac{1}{x}$? Using the weighted form of Jensen, $$\sum_{\text{cyc}} (\frac{a}{a+b+c+d} \cdot \frac{1}{b+2c+3d}) \geq \frac{1}{\frac{1}{a+b+c+d} \cdot (\sum_{\text{cyc}} a(b+2c+3d))} = \frac{a+b+c+d}{4(ab+ac+ad+bc+bd+cd)}$$$$\Rightarrow \sum_{\text{cyc}} \frac{a}{b+2c+3d} \geq \frac{(a+b+c+d)^2}{4(ab+ac+ad+bc+bd+cd)}$$and then proceed as in the above solutions. Thanks!

My Solution: So in this problem we can apply holder's inequality or we can say cauchy's functional equation since holder at power 1 is cauchy so yeah whatever. this can be written as\[\sum_{cyc}\frac{a}{b+2c+3d}\geq \frac23\]we can apply holder's here like below \[\left(\sum_{cyc}\frac{a}{b+2c+3d}\right)^1\left(\sum_{cyc}a(b+2c+3d)\right)^1\geq (a+b+c+d)^2\]So its suffice to prove that \[(a+b+c+d)^2\geq \frac23\left(\sum_{cyc}a(b+2c+3d)\right)\]\[a^2+b^2+c^2+d^2+2ab+2ac+2ad+2bc+2bd+2cd\geq \frac23\left(\sum_{cyc}ab+2ac+3ad\right)\]\[\implies \sum_{cyc}a^2+2\sum_{sym}ab\geq \frac23\left(4\sum_{sym}ab\right) \]\[\implies 3\sum_{cyc}a^2+6\sum_{sym}ab\geq 8\sum_{sym}ab\]\[\implies 3(a^2+b^2+c^2+d^2)\geq 2\sum_{sym}ab\]which is true by AM-GM so we are done !.$\blacksquare$

By C-S and AM-GM: $$\sum_{\text{cyc}}\frac a{b+2c+3d}=\sum_{\text{cyc}}\frac{a^2}{ab+2ac+3ad}\ge\frac{(a+b+c+d)^2}{4(ab+ac+ad+bc+bd+cd)}=\frac{a^2+b^2+c^2+d^2+2(ab+ac+ad+bc+bd+cd)}{4(ab+ac+ad+bc+bd+cd)}\ge\frac{\frac83}4=\frac23.$$

this inequalities homogeneous , so we can assume that $a+b+c+d=1$ and by titu's lemma we have $LHS \ge \dfrac{(a+b+c+d)^2}{4(ab+ad+ac+bc+dc+bd)}$ we know that $ab+ad+ac+bc+dc+bd\le \frac83$ (by newton's inequality) so $LHS\ge \frac23$

Let $a,b,c,d\geq 0$ and$(b+2c+3d)(c+2d+3a)(d+2a+3b)(a+2b+3c)\neq 0.$ Prove that $$\frac{ab}{(b+2c+3d)(c+2d+3a)} +\frac{cd}{(d+2a+3b)(a+2b+3c)} \leq \frac{1}{3} $$

By Titu: $$\sum_{\text{cyc}} \frac{a}{b+2c+3d}\geqslant \frac{(a+b+c+d)^2}{4(ab+ac+ad+bc+bd+cd)}$$It suffices to prove: $$ \frac{(a+b+c+d)^2}{4(ab+ac+ad+bc+bd+cd)} \geqslant \frac{2}{3}$$$$\iff 4(a^2+b^2+c^2+d^2)\geqslant (a+b+c+d)^2$$$$\iff (a-b)^2+(a-c)^2+(a-d)^2+(b-c)^2+(b-d)^2+(c-d)^2 \geqslant 0$$

$\color{magenta} \boxed{\textbf{SOLUTION A3}}$ By $\textbf{Titu's Inequality,}$ $$\frac{a^2}{a(b+2c+3d)} +\frac{b^2}{b(c+2d+3a)} +\frac{c^2}{c(d+2a+3b)}+ \frac{d^2}{d(a+2b+3c)} \geq \frac{(a+b+c+d)^2}{4(ab+ac+ad+bc+bd+cd)}$$$$\implies \frac{(a+b+c+d)^2}{4(ab+ac+ad+bc+bd+cd)}\ge \frac{2}{3}$$$$\implies 3(a+b+c+d)^2\ge 8(ab+ac+ad+bc+bd+cd)$$$$\implies 4(a^2+b^2+c^2+d^2) \ge (a+b+c+d)^2$$ But by $\textbf{Cauchy-Schwarz Inequality,}$ $$(1^2 +1^2 +1^2 + 1^2)(a^2+b^2+c^2+d^2) \ge (a+b+c+d)^2 \implies 4(a^2+b^2+c^2+d^2) \ge (a+b+c+d)^2 \blacksquare$$

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jawadkaleem wrote: What if we take a=b=c=d? we get 4/7>=2/3 which is not true. Can anyone help?? $1+2+3=6$.