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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+xzUsing 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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