Apply inequality cauchy, we have:

\(\left\{{}\begin{matrix}\left(b+c-a\right)+\left(a+c-b\right)\ge2\sqrt{\left(b+c-a\right)\left(a+c-b\right)}\\\left(a+c-b\right)+\left(a+b-c\right)\ge2\sqrt{\left(a+c-b\right)\left(a+b-c\right)}\\\left(a+b-c\right)+\left(b+c-a\right)\ge2\sqrt{\left(a+b-c\right)\left(b+c-a\right)}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}2c\ge2\sqrt{\left(b+c-a\right)\left(a+c-b\right)}\\2a\ge2\sqrt{\left(a+c-b\right)\left(a+b-c\right)}\\2b\ge2\sqrt{\left(a+b-c\right)\left(b+c-a\right)}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}c^2\ge\left(b+c-a\right)\left(a+c-b\right)\\a^2\ge\left(a+c-b\right)\left(a+b-c\right)\\b^2\ge\left(a+b-c\right)\left(b+c-a\right)\end{matrix}\right.\)

\(\Rightarrow\left(abc\right)^2\ge\left[\left(b+c-a\right)\left(a+c-b\right)\left(a+b-c\right)\right]^2\)

\(\Rightarrow abc\ge\left(b+c-a\right)\left(a+c-b\right)\left(a+b-c\right)\)

Apply inequality Cauchy, we have:

\(a+b\ge2\sqrt{ab}\)

\(\Rightarrow\left\{{}\begin{matrix}a^4+b^4\ge2\sqrt{a^4b^4}\\c^4+d^4\ge2\sqrt{c^4d^4}\end{matrix}\right.\)

\(\Leftrightarrow a^4+b^4+c^4+d^4\ge2a^2b^2+2c^2d^2=2\left[\left(ab\right)^2+\left(cd\right)^2\right]\)

But \(\left(ab\right)^2+\left(cd\right)^2\ge2\sqrt{\left(ab\right)^2\cdot\left(cd\right)^2}=2abcd\)

\(\Leftrightarrow a^4+b^4+c^4+d^4\ge2\cdot2abcd=4abcd\)

So \(a^4+b^4+c^4\ge abc\left(a+b+c\right)\)

We have:

\(a^2 +b^2+c^2\ge ab+bc+ca\)

Prove that it: 

This inequality <=> \(2a^2+2b^2+2c^2 \ge 2ab+2bc+2ca \)

<=> \(2a^2+2b^2+2c^2-2ab-2bc-2ca\ge0\)

<=> \((a-b)^2 + (b-c)^2 + (c-a)^2 \ge 0 \) (true) 

So a² + b² + c² \(\ge\) ab + bc + ca

Apply this for the topic, we have:

\(a^4+b^4+c^4 \ge (ab)^2 + (bc)^2 + (ca)^2 \)

\((ab)^2 + (bc)^2 + (ca)^2 \ge ab^2c + a^2bc + abc^2 = abc(a+b+c) \)

\(8(a^3 + b^3+c^3)\ge (a+b)^3 + (b+c)^3 + (c+a)^3 \)

\(\Leftrightarrow 8(a^3 +b^3+c^3) \ge 2(a^3+b^3+c^3) + 3ab(a+b) + 3bc(b+c)+3ca(c+a) \)

\(\Leftrightarrow\)\(6(a^3 +b^3+c^3) \ge 3[ab(a+b) + bc(b+c) + ca(c+a)] \)

\(\Leftrightarrow\) \(2a^3+2b^3+2c^3 \ge ab(a+b) + bc(b+c) + ca(c+a) \)

We have: a³ + b³ \(\ge \) a²b + ab²

Prove that the same way as the way of Kaya Renger to prove that: \(a^4+b^4 \ge a^3b+ab^3\)

So \(\left\{{}\begin{matrix}a^3+b^3\ge ab\left(a+b\right)\\b^3+c^3\ge bc\left(b+c\right)\\c^3+a^3\ge ca\left(c+a\right)\end{matrix}\right.\)

So true. 

=> \(8(a^3 + b^3+c^3)\ge (a+b)^3 + (b+c)^3 + (c+a)^3 \)

\(\left(a+b+c\right)^3\ge a^3+b^3+c^3+24abc\)

\(\Leftrightarrow a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge a^3+b^3+c^3+24abc\)

\(\Leftrightarrow3\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge24abc\)

\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge8abc\)

Apply inequality Cauchy, we have:

\(\left\{{}\begin{matrix}a+b\ge2\sqrt{ab}\\b+c\ge2\sqrt{bc}\\c+a\ge2\sqrt{ca}\end{matrix}\right.\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge8abc\)(True) 

So \(\left(a+b+c\right)^3\ge a^3+b^3+c^3+24abc\)

a+b+c = 0

=> (a+b+c)² = 0

=> a² + b² + c² + 2ab + 2bc + 2ca = 0

We have: a² + b² + c^2 \(\ge0\)

^{\(\Rightarrow2\left(ab+bc+ca\right)\le0\Leftrightarrow ab+bc+ca\le0\)}

Oh, sorry, the last row must be \(\ge-\left(2abc+2\right)\) not \(\ge-\left(abc+2\right)\)

\(a^2+b^2+c^2+2abc+1\ge2\left(ab+bc+ca\right)\)

\(\Leftrightarrow2\left(a^2+b^2+c^2+2abc+1\right)\ge2\left(ab+bc+ca\right)\)

\(\Leftrightarrow2a^2+2b^2+2c^2+4abc+2\ge2ab+2bc+2ca\)

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca\ge-4abc-2\)

\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge-\left(abc+2\right)\) (True because -(abc+2) \(\le0\) because a,b,c > 0) 

So .................

1st way: \(\sqrt{8}=\text{2.82842712475}\)

               \(\sqrt{5}+1=\text{3.2360679775}\)

So \(\sqrt{8}< \sqrt{5}+1\)

2nd way:

\(\sqrt{8}< \sqrt{9}=3\)

\(\sqrt{5}+1>\sqrt{4}+1=2+1=3\)

So \(\sqrt{8}< \sqrt{5}+1\)

Put B = 16ⁿ - 15n - 1 

* With n = 0 <=> B₀ = 16⁰ - 15.0 - 1 = 0 ⋮ 225 

* With n = 1 <=> B₁ = 16¹ - 15.1 - 1 = 0 ⋮ 225

..........................

* With n = k <=> B_k = 16^k - 15k - 1 ⋮ 225

We must prove that with n = k+1 then B ⋮ 225 too

We have: n = k+1

B_k+1 = 16^k+1 - 15.(k+1) - 1 = 16^k.16 - 15k - 15 - 1 = 16^k.16 - 15k - 16 

= 16(16^k - 1) - 15k  

So B_k+1 - 225k = 16(16^k - 15k - 1) = 16.B_k

B_k ⋮ 225, 225 ⋮ 225 so B_k+1 ⋮ 225.

So 16ⁿ - 15n - 1 ⋮ 225  

34² + 66² + 68.66

= 34² + 2.34.66 + 66² 

= (34 + 66)² = 100² = 10000 

Multiplying each side with 2, we have:

2xy + 2yz + 2zx ≤ 2x² + 2y² + 2z²

<=> 2xy + 2yz + 2zx - 2x² - 2y² - 2z² ≤ 0

<=> 2xy + 2yz + 2zx - x² - x² - y² - y² - z² - z² ≤ 0

<=> -(x² - 2xy + y²) - (x² - 2xz + z²) - (y² + 2yz + z²) ≤ 0

<=> -(x+y)² - (x+z)² - (y+z)² ≤ 0 (True)     

So xy + yz + zx ≤ x² + y² + z²

9999 => 8765 => 7531

We have:

CD.DA = 24; \(\dfrac{FD\cdot DA}{2}=9\Rightarrow FD\cdot DA=18\)

=> \(\dfrac{FD}{CD}=\dfrac{18}{24}=\dfrac{3}{4}\Rightarrow FC=\dfrac{1}{4}CD\)

\(\dfrac{AB\cdot BE}{2}=4\Rightarrow AB\cdot BE=8\)

\(\Rightarrow\dfrac{BE}{BC}=\dfrac{8}{24}=\dfrac{1}{3}\Rightarrow EC=\dfrac{2}{3}BC\)

\(\Rightarrow S_{\Delta EFC}=\dfrac{\dfrac{1}{4}CD\cdot\dfrac{2}{3}BC}{2}=\dfrac{\dfrac{1}{6}.24}{2}=\dfrac{4}{2}=2\left(cm^2\right)\)

\(\Rightarrow S_{\Delta AEF}=24-9-4-2=13\left(cm^2\right)\)

202 + ⊠⊠⊠ + ⊠⊠⊠ = 2002

=> ⊠⊠⊠ + ⊠⊠⊠ = 2002 - 202 = 1800

We have: 819 + 981 = 1800; 849 + 951 = 1800;.........

So the sum of all the 6 missing digits are: 8 + 1 + 9 + 9 + 8 + 1 = 8 + 4 + 9 + 9 + 5 + 1 = ... = 36

Wrong post

Because if a,b,c > 0 so

\(\left\{{}\begin{matrix}a\ge\sqrt{a}\\b\ge\sqrt{b}\\c\ge\sqrt{c}\end{matrix}\right.\Rightarrow a+b+c\ge\sqrt{a}+\sqrt{b}+\sqrt{c}\)

a, \(f\left(2\right)=2^3-4.2^2-7.2-1=8-16-14-1=-23\)

\(g\left(\dfrac{1}{2}\right)=\left(\dfrac{1}{2}\right)^4+\left(\dfrac{1}{2}\right)^3-\left(\dfrac{1}{2}\right)^2-7.\dfrac{1}{2}+2=\dfrac{1}{16}+\dfrac{1}{8}-\dfrac{1}{4}-\dfrac{7}{2}+2=-\dfrac{25}{16}\)

b,

\(f\left(x\right)-g\left(x\right)=x^3-4x^2-7x-1-\left(x^4+x^3-x^2-7x+2\right)\)

\(=x^3-4x^2-7x-1-x^4-x^3+x^2+7x-2\)

\(=-x^4-3x^2-3\)

AB = AC = 8 => \(\Delta ABC\) is a isosceles triangle

=> AM is high road, median (trung tuyến) of this triangle.

=> AM \(\perp\) BC and MB = MC = 5

=> AM = \(\sqrt{8^2-5^2}=\sqrt{39}\)

Tell the score to find is x

\(\Rightarrow\dfrac{82+86+92+x}{4}=90\)

=> 82 + 86 + 92 + x = 360

=> 260 + x = 360

=> x = 100

So she needs on her fourth test to have the average is 90