8

Putting everything over a common denominator shows that $$I(a,b) = \frac{a^2+b^2+a^2b+ab^2+a+b}{a^2+b^2+a^2b+ab^2+a+b+2ab} = \frac{(1+a)(1+b)(a+b)-2ab}{(1+a)(1+b)(a+b)} \leq 1$$ since $a,b\geq 0$. Simplifying this expression, we get $$I(a,b) = 1 - \frac{2ab}{(1+a)(1+b)(a+b)}$$ Now take a look at the second term and consider it rewritten in the following ...


6

$$\sum_{cyc}\frac{a}{b^2+1}=3+\sum_{cyc}\left(\frac{a}{b^2+1}-a\right)=3-\sum_{cyc}\frac{ab^2}{b^2+1}\geq$$ $$\geq3-\sum_{cyc}\frac{ab^2}{2b}=3-\frac{1}{2}(ab+ac+bc).$$ Can you end it now? Since by your work $$3-\frac{1}{2}(ab+ac+bc)=3-\frac{1}{2}\cdot\frac{9-a^2-b^2-c^2}{2},$$ it's enough to prove that $$3-\frac{1}{2}\cdot\frac{9-a^2-b^2-c^2}{2}\geq\frac{3}...


6

You need to use another queue: By Rearrangement, AM-GM and C-S we obtain: $$\sum_{i=1}^n\left(\frac{a_i^3+1}{a_i^2+1}\right)^4\geq \ \sum_{i=1}^n\left(\frac{a_i+1}{2}\right)^4\geq\sum_{i=1}^na_i^2=\frac{1}{n}\sum_{i=1}^n1^2\sum_{i=1}^na_i^2\geq\frac{1}{n}\left(\sum_{i=1}^na_i\right)^2=n. $$ I used the following. $$\frac{x^3+1}{x^2+1}\geq\frac{x+1}{2}$$ it'...


5

While I agree with the comments that this is an art rather than a science, there are patterns, and one important one (at least for me) came from The Cauchy-Schwarz Master Class (J. Michael Steele). You typically get a "good deal" from inequalities when you're working in a region where the inequality is sharp. As an (obvious) example, say you're working ...


5

By Show that $\frac{x}{1-x}\cdot\frac{y}{1-y}\cdot\frac{z}{1-z} \ge 8$. $$\frac{a}{1-a}\cdot\frac{b}{1-b}\cdot\frac{c}{1-c}\geq 8$$ where the equality holds for $a=b=c=2/3$. Moreover, for $t\geq 3$, by letting $b(t)=c(t)=1-1/t\in (0,1)$ and $a(t)=2/t\in (0,1)$, then $a(t)+b(t)+c(t)=2$ and $$g(t):=\frac{a(t)}{1-a(t)}\cdot\frac{b(t)}{1-b(t)}\cdot\frac{c(t)}{...


5

Let $y_k = \dfrac{m}{x_k + m}, 0 < y_k <1, \sum{y_k}=1$. Since $x_k=m \cdot \left(\dfrac{1-y_k}{y_k}\right)$, the inequality to be proven translates as $\prod{(1-y_k)}\ge (n-1)^n\prod{y_k}$. Noting that $1-y_k =\sum {y_q}-y_k$, applying the mean inequality gives $(1-y_k)^{n-1} \ge (n-1)^{n-1}\dfrac{\prod {y_q}}{y_k}$ for eack $k=\overline{1, n}$. ...


5

Basically you're being asked to show that $d \leq {3abc \over ab + bc + ac}$ implies $d \leq {ab + bc + ac \over a + b + c}$. This is equivalent to showing that for all positive $a,b,c$ you have $${3abc \over ab + bc + ac} \leq {ab + bc + ac \over a + b + c}$$ Let $a = {1 \over x}$, $b = {1 \over y}$, and $c = {1 \over z}$. Then what you are trying to show ...


4

Let $y_i:=\dfrac{1}{1+x_i}$ for $i=0,1,2,\ldots,n$. Then, $y_0,y_1,y_2,\ldots,y_n$ are positive real numbers such that $\sum\limits_{i=0}^n\,y_i\leq 1$. We want to prove that $$\prod_{i=0}^n\,\left(\frac{1-y_i}{y_i}\right)\geq {n^{n+1}}.$$ Let $[n]:=\{0,1,2,\ldots,n\}$. To prove the last inequality, we note that $$1-y_i\geq \sum_{j\in[n]\setminus\{i\}}\,...


4

We use the inequality $$\frac{a+b}{2}\le \sqrt{\frac{a^2+b^2}{2}}$$ so we get $$\frac{1+\sqrt{2}}{2}\times\frac{1+\sqrt{3}}{2}\times\frac{1+\sqrt{5}}{2}\le \sqrt{30}$$


4

Let $x\geq y\geq z$. Thus, $x\geq34$ and $z\leq33$ and since $$x+y=100-z\geq100-33=34+33,$$ we obtain $$(x,y,z)\succ(34,33,33)$$ and since $f(x)=e^x$ increases and $\ln$ is a concave function, by Karamata we obtain: $$xyz=e^{\ln{x}+\ln{y}+\ln{z}}\leq e^{\ln34+\ln33+\ln33}=34\cdot33^2$$ and we are done!


4

A Hint for an Alternative Solution. We want to show that $$\left(\frac{x^3+1}{x^2+1}\right)^4\geq 2x-1$$ for every $x\in\mathbb{R}$. By the AM-GM Inequality, $$\left(\frac{x^3+1}{x^2+1}\right)^4+1\geq 2\,\left(\frac{x^3+1}{x^2+1}\right)^2\,.$$ Hence, it suffices to verify that $$\left(\frac{x^3+1}{x^2+1}\right)^2\geq x$$ for all $x\in\mathbb{R}$. This ...


3

I think, it means that $a+b>0$, otherwise your inequality is wrong for $a=b=0$. Let $a+b=2x$. Thus, by C-S and AM-GM we obtain: $$I(a,b)=\frac{1}{a+b}\left(\frac{a^2}{1+a}+\frac{b^2}{1+b}\right)+\frac{1}{1+a+b+ab}\geq$$ $$\geq\frac{1}{2x}\cdot\frac{(a+b)^2}{1+a+1+b}+\frac{1}{1+2x+\left(\frac{a+b}{2}\right)^2}=$$ $$=\frac{x}{1+x}+\frac{1}{(x+1)^2}=\frac{...


3

By AM-GM $$a^2+ab+b^2=\sqrt[3]{\frac{1}{8}(2a^2+2ab+2b^2)^3}=\frac{1}{2}\sqrt[3]{((a+b)^2+a^2+b^2)^3}\geq$$ $$\geq\frac{1}{2}\sqrt[3]{\left(2\cdot\frac{(a+b)^2}{2}+2ab\right)^3}\geq\frac{1}{2}\sqrt[3]{\left(3\sqrt[3]{\left(\frac{(a+b)^2}{2}\right)^2\cdot2ab}\right)^3}=$$ $$=\frac{3}{2}\sqrt[3]{\frac{(a+b)^4ab}{2}}\geq\frac{3}{2}\sqrt[3]{\frac{(a+b)^2\cdot4ab\...


2

It turns out that this question is non-trivial and was resolved only in 1999, by Rajendra Bhatia and Fuad Kittaneh in the paper Notes on matrix arithmetic–geometric mean inequalities, freely available here. They show that, for any unitarily invariant matrix norm $\|.\|$, $$ \| A B\|\leq \frac 14 \|(A+B)^2\|. $$ In the particular case of the Schatten norm ...


2

Since for $x>0$ we have (just draw a graph for ${1\over 1+x^2}$ and a tangent at $x=1$) $${1\over 1+x^2}\geq -{1\over 2}x+1$$ it is enough to check if $$-{1\over 2}(ab+bc+ca)+3\geq {3\over 2}$$ i.e. $$3\geq ab+bc+ca$$ is true? Since $$a^2+b^2+c^2\geq ab+bc+ca$$ that is easy to verfy. :)


2

Let $\frac{a}{1-a}=x$, $\frac{b}{1-b}=y$ and $\frac{c}{1-c}=z.$ Thus, $x$, $y$ and $z$ are positives, $a=\frac{x}{1+x},$ $b=\frac{y}{1+y},$ $c=\frac{z}{1+z}$ and the condition gives: $$\sum_{cyc}\frac{x}{1+x}=2.$$ Id est, by AM-GM we obtain: $$\prod_{cyc}\frac{x}{1+x}=\prod_{cyc}\left(2-\frac{y}{1+y}-\frac{z}{1+z}\right)=\prod_{cyc}\left(\frac{1}{1+y}+\...


2

In your AM-GM the equality occurs for $$2-2x=2-y=2x+2y=\frac{1}{4},$$ which is impossible. I think, it's better to make the following. By AM-GM $$(1-x)(2-y)^2(x+y)=4(1-x)\left(1-\frac{y}{2}\right)^2(x+y)\leq$$ $$\leq4\left(\frac{1-x+2\left(1-\frac{y}{2}\right)+x+y}{4}\right)^4=\frac{81}{64}.$$ The equality occurs for $$1-x=1-\frac{y}{2}=x+y$$ or $$(x,y)=\...


2

By C-S $$\sum_{cyc}\frac{a}{b+1}=\sum_{cyc}\frac{a^2}{ab+a}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}(ab+a)}.$$ Thus, it's enough to prove that $$2(a+b+c)^2\geq3\sum_{cyc}(ab+a)$$ or $$\sum_{cyc}(2a^2+ab)\geq3(a+b+c).$$ Now, by AM-GM $$\sum_{cyc}ab\geq3\sqrt[3]{a^2b^2c^2}=3$$ and by C-S again $$a^2+b^2+c^2=\frac{1}{3}(1+1+1)(a^2+b^2+c^2)\geq\frac{1}{3}(a+b+c)^2.$...


2

By adding $3$ to both sides of the given inequality, it is equivalent to $$\frac{(a+k)+(b+k)+(c+k)}{k}\geq \frac{a+k}{b+k}+\frac{b+k}{c+k}+\frac{c+k}{a+k}.\ \ \ \ \ (1)$$ This is also equivalent to $$\frac{A+B+C}{k}\ge \frac{A}{B}+\frac{B}{C}+\frac{C}{A}\ \ \ \ \ (2)$$ for all $A,B,C\geq k$ and $k>0$ (by setting $A=a+k$, $B=b+k$, and $C=c+k$). But this ...


2

$$\sum_{k\neq i}\frac{1}{x_k+m}=\frac{1}{m}-\frac{1}{x_i+m}=\frac{x_i}{m(x_i+m)}.$$ Thus, by AM-GM $$\prod_{i=1}^n\frac{x_i}{x_i+m}=m^n\prod_{i=1}^n\sum_{k\neq i}\frac{1}{x_k+m}\geq\frac{m^n(n-1)^n}{\prod\limits_{i=1}^n\left(\prod\limits_{k\neq i}(x_k+m)\right)^{\frac{1}{n-1}}}=\frac{m^n(n-1)^n}{\prod\limits_{i=1}^n(x_i+m)},$$ which ends a proof.


2

We have $$a^2+b^2+c^2+2ab+2bc+2ac=(a+b+c)^2=1$$ which proves your inequality. Supposing that $a,b,c\neq 0$ we see that equality in the weighted AM-GM occurs if and only if $a=b=c=\frac13$.


2

The inequality holds if and only if $3a^{2}(2-a)+3(1-a)^{2}(a+1)-(a+1)(2-a)\geq 0$, the former is just $4a^{2}-4a+1$ which is just $(2a-1)^{2}$.


2

By C-S $$\frac{a^2}{a+1}+\frac{(1-a)^2}{2-a}\geq\frac{(a+1-a)^2}{a+1+2-a}=\frac{1}{3}.$$ I used the following C-S: For any reals $a_i$ and positives $b_i$ we have: $$\frac{a_1^2}{b_1}+\frac{a_2^2}{b_2}+...+\frac{a_n^2}{b_n}\geq\frac{(a_1+a_2+...+a_n)^2}{b_1+b_2+...+b_n}.$$


2

By AM-GM: $$\sum_{cyc}\frac{x}{x+y+z}\left(1+\frac{y-z}{x}\right)\geq\prod_{cyc}\left(1+\frac{y-z}{x}\right)^{\frac{x}{x+y+z}},$$ which gives $$1\geq\prod_{cyc}\left(1+\frac{y-z}{x}\right)^{\frac{x}{x+y+z}}$$ or $$1\geq\prod_{cyc}\left(1+\frac{y-z}{x}\right)^x$$ and we got a reversed inequality.


2

Cauchy-Schwarz ineuality implies that $$\left(\frac{x^2}{x+z} + \frac{y^2}{y+x} + \frac{z^2}{z+y}\right)\left((x+z)+(y+x)+(z+y)\right)\geq(x+y+z)^2,$$ i.e. $$\left(\frac{x^2}{x+z} + \frac{y^2}{y+x} + \frac{z^2}{z+y}\right)\times 2\geq 1,$$ which gives the result.


1

By weighted AM-GM we can wtite $$(a^a~b^c~c^c)^{(a+b+c)^{-1}} \le \frac{a.a+b.b+c.c}{a+b+c}~~~(1)$$ and $$(a^b~b^c~c^a)^{(a+b+c)^{-1}} \le \frac{a.b+b.c+c.a}{a+b+c}~~~(2)$$ and $$(a^c~b^a~c^b)^{(a+b+c)^{-1}} \le \frac{a.c+b.a+c.b}{a+b+c}~~~(3)$$ Adding all these equations and using $a+b+c=1$. we get $$a^ab^bc^c+a^bb^cc^a+a^cb^ac^b \le (a^2+b^2+c^2+2(ab+bc+...


1

Because by AM-GM $$\prod_{i=0}^n\frac{x_i}{1+x_i}\geq\prod_{i=0}^n\sum_{k\neq i}\frac{1}{1+x_k}\geq\prod_{i=0}^n\left(n\prod_{k\neq i}\frac{1}{\sqrt[n]{1+x_k}}\right)=\frac{n^{n+1}}{\prod\limits_{i=0}^n(1+x_i)}$$ and we are done!


1

A much more Logical way to solve this without algebraic manipulations: Since we want to maximize $(1-x)(2-y)^2(x+y)$ , and all the terms are positive, we can see that we might want to maximize $(2-y)^2$ in particular as it is the biggest among all terms. $$\max \space(2-y)^2 = (2-0)^2 = 4 \space\space\space\space\space\space\space \text for \space y = 0$$ ...


1

As I have written, the idea of proof taking $c=mid(a,b,c)$ is just a matter of notation. Without loss of generality (WLOG), You can take the other mid one if $c$ is not the one. Note that there is always something in the middle. Even if $a=b=c$ you can take any of them.


1

After squaring of the both sides we need to prove that $$\sum_{sym}(a^4b^2-a^4bc+a^3b^3-2a^3b^2c+a^2b^2c^2)\geq0,$$ which is true by Muirhead and Schur.


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