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\(\dfrac{a^2+b^2}{2}\ge\dfrac{\left(a+b\right)^2}{2^2}\)

<=> \(4\left(a^2+b^2\right)\ge2\left(a+b\right)^2\)

<=> \(4\left(a^2+b^2\right)\ge2\left(a^2+2ab+b^2\right)\)

<=>  \(\left(2a^2+2b^2-4ab\right)\ge0\)

<=> \(2\left(a-b\right)^2\ge0\)   (Right)

So .... 

1^st way : We have

\(\sqrt{8}^2=8=6+2\)

\(\left(\sqrt{5}+1\right)^2=5+2\sqrt{5}+1=6+2\sqrt{5}\)

Cause \(2\sqrt{5}\ge2\sqrt{4}=4>2\)

So \(\sqrt{8}< \sqrt{5}+1\)

2^st way :

\(\sqrt{8}< \sqrt{9}=3=2+1=\sqrt{4}+1< \sqrt{5}+1\)

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 \(\sqrt{8}< \sqrt{5}+1\)

\(\dfrac{a}{ab+a+2}+\dfrac{b}{bc+b+1}+\dfrac{2c}{ac+2c+2}=\dfrac{ac}{2+ac+2c}+\dfrac{ac.b}{abc.c+abc+ac}+\dfrac{2c}{ac+2c+2}\)

                                                                          \(=\dfrac{ac}{2+ac+2c}+\dfrac{2}{2c+2+ac}+\dfrac{2c}{ac+2c+2}=1\)

Because the role of a,y,z is the same so we suppose \(x\le y\le z\)

\(2^z< 2336\Rightarrow z\le11\)

We have:\(2^x+2^y+2^z\le3.2^z\Leftrightarrow2336\le3.2^z\)\(\Leftrightarrow2^z\ge\dfrac{2336}{3}\Rightarrow z\ge10\)

So z\(\in\left\{10,11\right\}\)

With z=10 then \(2^x+2^y=2336-2^{10}=1312\)

    \(2^x+2^y\le2.2^y\Rightarrow2.2^y\ge1312\Rightarrow2^y\ge656\Rightarrow y\ge10\)

  \(2^y< 1312\Rightarrow y\le10\)

 So y=10 so \(2^x=2336-2^{10}-2^{10}=288\)(not satisfy)

With z=11 then \(2^x+2^y=288\)

 \(2^y< 288\Rightarrow y\le8\)

 \(2^y.2\ge2^x+2^y=288\Rightarrow2^y\ge144\Rightarrow y\ge8\)

So y=8 then \(2^x=288-2^8=32\Rightarrow x=5\)

Answer :(x,y,z) is the permutation of \(\left(5,8,11\right)\)

oh;the first row :x,y,z not a,y,z

34² + 66² + 68.66

= 34² + 2.34.66 + 66² 

= (34 + 66)² = 100² = 10000 

34² + 66² + 68 . 66

= (34² + 66² + 2 . 34 . 66)

= (34 + 36)² 

=100²

=> 10000

\(34^2+66^2+68.66=\left(34^2+66^2+2.34.66\right)=\left(34+66\right)^2=100^2=10000\)

+) With n = 1 then 

\(16^n-15n-1=16-15-1=0⋮225\)

+) With n = 2 then 

\(16^n-15n-1=16^2-15.2-1=225⋮225\)

Suppose \(\left(16^n-15n-1\right)⋮225\)

We are going to prove that \(\left(16^{n+1}-15\left(n+1\right)-1\right)⋮225\) too

We can see \(16^{n+1}-15\left(n+1\right)-1=16.16^n-15n-15-1=\left(16^n-15n-1\right)+15.16^n-15\)

Following Inductive hypothesis , \(\left(16^n-15n-1\right)⋮225\)

And \(15.16^k-15=15\left(16^k-1\right)⋮15.15=225\)

So \(\left(16^{n+1}-15\left(n+1\right)-1\right)⋮225\)

Conclude \(\left(16^n-15n-1\right)⋮225\) \(\forall n\in N\)

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  

With n=0 then \(2^0+1=1+1=2⋮̸3\)(not satisfy)

With n=1 then \(2^{2.1}+1=2+1=3⋮3\)(satisfy)

With n>1 then \(2^{2n}=\left(2^2\right)^n=4^n\equiv1^n=1\)(mod 3)\(\Rightarrow2^{2n}+1\equiv2\)(mod 3)

\(\Rightarrow2^{2n}+1⋮̸3\)(not satisfy)

Answer: n=1