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12

The solution follows from the following lemma. Lemma. For $z\in\mathbb C$, we have that $$|z^2+1|<1\qquad\text{implies}\qquad |z+1|>\frac{1}{\sqrt 2}.$$ The solution follows easily once we have the lemma, since we may introduce a positive real $r>0$ and replace $z$ with $z/r$ to deduce that $$ |z^2+r^2|<r^2\qquad\text{implies}\qquad |z+r|>\...


7

There is an error; the factorisation $x^{1/n}(x-1) - x(x-1)^{1/n} = x^{1/n} (x-1)^{1/n} ((x-1)-x)$ is incorrect and should be $x^{1/n}(x-1) - x(x-1)^{1/n} = x^{1/n} (x-1)^{1/n} ((x-1)^\frac{n-1}{n}-x^\frac{n-1}{n})$. To correctly prove this, note that since $\frac{1}{n}-1$ is negative we have $x^{\frac{1}{n}-1}$ is decreasing. This implies $(x-1)^{\frac{1}{...


7

Since squaring is monotonic, this is equivalent to $$ a^2 + b^2 + c^2 + d^2 + 2\sqrt{a^2+c^2}\sqrt{b^2+d^2} \ge a^2+b^2+c^2+d^2 + 2(ab + cd) $$ which is in turn equivalent to $$ ab + cd \le \sqrt{a^2+c^2}\sqrt{b^2+d^2}, $$ the Cauchy-Schwarz inequality. Putting it in vector form is even more elegant. For $\mathbf{a} = (a,c)$ and $\mathbf{b} = (b,d)$, this ...


7

Drafting behind Michael Rozenberg's clever answer, appealing to the concavity of $\sin$ on $[0, \pi]$ quickly reduces the problem to showing the inequality $$2 \sin 1 > \frac{8}{5} .$$ From $\pi < \frac{22}{7}$ we deduce $\frac{3 \pi}{10} < \frac{66}{70} < 1$, and so $$2 \sin 1 > 2 \sin \frac{3 \pi}{10} = 2 \cdot \frac{1}{4}(1 + \sqrt{5}) > ...


6

This can be written as $$\sqrt{2x}\geq\ln(1+x+\sqrt{2x}), \quad x\geq0$$ raising both sides to the exponential function, the relation becomes $$\text{e}^{\sqrt{2x}}\geq1+x+\sqrt{2x}$$ using the taylor expansion for the exponential function (and specifically writing out terms which will cancel with those on the right side), we have $$1+\sqrt{2x}+\frac{...


6

Both results are correct, but one of them tells you more than the other. For instance, if $x=20$, then both $x\ge10$ and $x\ge 5$ are correct, but $x\ge 10$ tells you more. In your case, the second approach tells you more, because it uses explicitly the knowledge that the first term is equal to $1$. In fact, with a bit of work, you could even use this to ...


6

Since by PM $$\left(\frac{x^n+1}{2}\right)^k\geq\left(\frac{x^k+1}{2}\right)^n$$ is true for all $x\geq0$ and $n\geq k>0,$ it's enough to prove that $$\frac{x^{m+1}+1}{x^m+1}\geq\sqrt[2m+1]{\frac{x^{2m+1}+1}{2}}$$ or $f(x)\geq0,$ where $$f(x)=\ln\left(x^{m+1}+1\right)-\ln\left(x^m+1\right)-\frac{1}{2m+1}\ln\left(x^{2m+1}+1\right)+\frac{\ln2}{2m+1}.$$ ...


6

$$\frac{xy}{z^2(x+y)}+\frac{yz}{x^2(z+y)}+\frac{xz}{y^2(x+z)}$$ $$=\frac{2x^2y^2}{z(x+y)}+\frac{2y^2z^2}{x(z+y)}+\frac{2x^2z^2}{y(x+z)}$$ $$\geq 2\frac{(xy+yz+xz)^2}{2(xy+yz+zx)}=xy+yz+xz$$ Using Titu's Lemma which is a variant of the Cauchy-Schwarz inequality.


6

Note that\begin{align}(\forall z\in\mathbb C):\bigl\lvert\cos(z)\bigr\rvert&=\left\lvert1-\frac{z^2}{2!}+\frac{z^4}{4!}-\frac{z^6}{6!}+\cdots\right\rvert\\&\leqslant1+\frac{\lvert z\rvert^2}{2!}+\frac{\lvert z\rvert^4}{4!}+\frac{\lvert z\rvert^6}{6!}+\cdots\\&\leqslant1+\lvert z\rvert+\frac{\lvert z\rvert^2}{2!}+\frac{\lvert z\rvert^3}{3!}+\frac{\...


6

It is $$\frac{a}{b}+\frac{b}{a}+\frac{b}{c}+\frac{c}{b}+\frac{c}{a}+\frac{a}{c}\geq 6$$ and now use that $$x+\frac{1}{x}\geq 2$$ for all $$x>0$$ You can also use that $$\frac{a^2 b+ab^2+b^2c+c^2b+a^2c+ac^2}{6}\geq \sqrt[6]{(abc)^6}=abc$$


5

By the weighted AM-GM inequality, $a^ab^bc^c \leq a \times a + b \times b + c \times c$ (by using weights $a$, $b$, $c$). Similar expressions hold for the other terms. So the inequality becomes $a^ab^bc^c + a^bb^cc^a + a^cb^ac^b \leq a^2 + b^2 + c^2 + ab + bc + ca + ac + ba + cb = (a+b+c)^2 = 1.$


5

$$0.99<0.9999^{100}$$ it's $$1-\frac{1}{100}<\left(1-\frac{1}{100^2}\right)^{100},$$ which is true by Bernoulli: $$\left(1-\frac{1}{100^2}\right)^{100}>1-100\cdot\frac{1}{100^2}=1-\frac{1}{100}.$$ $$0.9999^{101}<0.99$$ it's $$\left(1-\frac{1}{100^2}\right)^{101}<1-\frac{1}{100}$$ or $$\left(1-\frac{1}{100^2}\right)^{100}\left(1+\frac{1}{100}\...


5

Since $$(a_1+a_2+\cdots +a_n)^2=(a_1^2+a_2^2+\cdots +a_n^2)+2\sum_{1 \le i < j \le n} a_{i} a_{j}$$ we can write $$\sum_{1 \le i < j \le n} a_{i} a_{j}=\frac{S_n^2-(a_1^2+a_2^2+\cdots +a_n^2)}{2}$$ So, the hint $$ \sum_{1 \le i < j \le n} a_{i} a_{j} \ge \frac{n(n-1)}{2} $$ is equivalent to $$\frac{S_n^2-(a_1^2+a_2^2+\cdots +a_n^2)}{2}\ge \frac{n(n-...


5

Let $x\in [-1/2,1/2].$ By the MVT we have $$\ln (1-x) = \ln (1) -\frac{1}{1-c_x}x$$ for some $c_x$ between $0$ and $x.$ Thus $$|\ln (1-x)| = \left|\frac{1}{1-c_x}x\right|\le 2|x|.$$ We get the factor of $2$ on the right because $|c_x|\le 1/2,$ hence $1/(1-c_x)\le 1/(1-1/2)=2.$


5

Hint for one part, using this inequality and this one $$\cos{x} \geq 1 - \frac{x^2}{2}$$ we have $$\int\limits_{0}^{\frac{\pi}{2}}\cos{\sin{x}} dx > \int\limits_{0}^{\frac{\pi}{2}} \left(1-\frac{\sin^2{x}}{2}\right)dx=\frac{3 \pi}{8}$$


5

We need to prove that $$\sum_{cyc}(x^3+3x^2y+3x^2z+2xyz)+\sum_{cyc}3xyz\geq 4\sum_{cyc}(x^2y+x^2z+xyz)$$ or $$\sum_{cyc}(x^3-x^2y-x^2z+xyz)\geq0,$$ which is Schur. Your way is wrong because you took too strong estimation, that got a wrong inequality. Here happens like the following. Let we need to prove that $2>1$. We know that $1<3$, but it does ...


5

$$\left\vert\frac{e^{-ixu}-1}{u}\right\vert^2=\frac{2-2\cos(xu)}{u^2}=\frac{4\sin^2(\frac{xu}{2})}{u^2}\leq\frac{4(\frac{xu}{2})^2}{u^2}=x^2$$


5

Yes, it is true. Jensen's inequality was applied in the last step $$\log \left( \frac{1}{n} \sum \limits_{i=1}^{n}{\alpha_i} \right) \geq \frac{1}{n} \sum _{i=1}^n {\log(\alpha_i)},$$ since $\log$ is concave.


5

Let $p+1$, $q+1$, $r+1$ be our roots. Thus, $p$, $q$ and $r$ are positives and by using the Viete's theorem we need to prove that: $$\sum_{cyc}(p+1)(q+1)+\prod_{cyc}(p+1)\geq3\sum_{cyc}(p+1)-5$$ or $$pqr+2(pq+pr+qr)\geq0,$$ which is obvious.


5

We have $w_0\le -w_1-w_2<2w_0$ with $w_0\le 0$. This is impossible.


4

What happens is that a square is always $\geq 0$. What I would do is split in 2 cases (when $x$ is positive or negative) If $0\leq x\leq 3$, then $0\leq x^2\leq 9$. If $-2\leq x\leq 0$, then $0\leq x^2\leq 4$ (think about it). So, by taking the "union of the sets" $0\leq x^2\leq 9$ and $0\leq x^2\leq 4$, you get all the possible values of $x^2$. You get ...


4

Remember that $a\leq b \iff a^2\leq b^2$ only if $a,b\geq 0$. If $x\geq 0$ then from $x\leq 3$ after squaring, we get $x^2\leq 9$ and if $x\leq 0$ then from $-2\leq x$ thus $-x\leq 2$ so after squaring we get $x^2\leq 4$. So $$0\leq x^2\leq 9$$


4

Try to use weighted AM-GM inequality: for any $x,y,z,p,q,r>0$ with $p+q+r=1$ one has $$x^py^qz^r \le px+qy+rz.$$


4

take the expression $$5x^{2}-2xy-8x+2y^{2}-2y+5$$, we want to write this as a sum of squares somehow. grouping the terms as: $$(x^{2}-2xy+y^{2}) + (y^{2}-2y+1) + (4x^{2}-8x+4) $$ factoring the three expressions gives $$(x-y)^{2} + (y-1)^{2} + 4(x-1)^{2}$$ which is the sum of squares, each of which is greater than $0$. for equality to hold, notice that both $(...


4

Bit late to the party. A sort of minimalist expression is $$ \frac{1}{5}(5x-y-4)^2 + \frac{9}{5} (y-1)^2 $$ so we get zero only when $y=1$ and $5x-1-4 =0,$ so also $x=1.$ $$ Q^T D Q = H $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ - \frac{ 1 }{ 5 } & 1 & 0 \\ - \frac{ 4 }{ 5 } & - 1 & 1 \\ \end{array} \right) \left( ...


4

Sum up $a_i^2+a_j^2 \geq 2a_i a_j$ for all pairs ($i,j$). You'll get, $$(n-1)\sum_ia_i^2 \geq 2\sum_{i<j}a_ia_j$$ Implying that, $$\left(\sum_i a_i\right)^2 = \sum_i a_i^2 + 2\sum_{i<j}a_ia_j \geq \left(2+\frac{2}{n-1}\right)\sum_{i<j}a_ia_j=\frac{2n}{n-1}\sum_{i<j}a_ia_j\geq n^2$$ $$\implies \sum_ia_i \geq n$$


4

It's $$(7^x-5^x)(4^x+3^x-5^x)\geq0.$$ Can you end it now? I got $0\leq x\leq2.$


4

So you should now prove that $$2\sqrt{n}+\frac{1}{\sqrt{n+1}}<2\sqrt{n+1}$$ by applying high-school algebra.


4

Drafting behind Michael Rozenberg's clever answer, appealing to the concavity of $\sin$ on $[0, \pi]$ quickly reduces the problem to showing the inequality $$2 \sin 1 > \frac{8}{5} .$$ Then, from Maclaurin expansion, we have $$ \sin 1 = 1 - \frac{1}{3!} + \frac{1}{5!} - \frac{1}{7!} + \ldots $$ Observe that the absolute value of each of these terms is ...


4

As Erick Wong says, we can WLOG that exactly two of the $x_i$ are negative, say $x_5,x_6$. Let $S=x_1+x_2+x_3+x_4$. Then by RMS-AM, $$\sum\limits_{i=1}^4 x_i^2\geq \frac{S^2}{4}$$ and similarly $$x_5^2+x_6^2\geq \frac{S^2}{2}$$ so $S\leq 2\sqrt{2}$. Since $\sqrt[4]{x_1x_2x_3x_4}\leq\frac{S}{4}$ and $\sqrt{x_5x_6}\leq\frac{S}{2}$, we know that $$\prod x_i\...


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