1)Were Mr.Morgan to be still head master,he wouldn't permit such bad behaviour.

2) Had anything gone wrong with my plan,I would have held responsibility.

Ta có:\(\left(a^2+bc\right)\left(b+c\right)=b\left(a^2+c^2\right)+c\left(a^2+b^2\right)\)

\(\Rightarrow\sqrt{\frac{\left(a^2+bc\right)\left(b+c\right)}{a\left(b^2+c^2\right)}}=\sqrt{\frac{b\left(a^2+c^2\right)+c\left(a^2+b^2\right)}{a\left(b^2+c^2\right)}}\)

Tương tự\(\Rightarrow\)VT=\(\Sigma\sqrt{\frac{b\left(a^2+c^2\right)+c\left(a^2+b^2\right)}{a\left(b^2+c^2\right)}}\)

Đặt \(x=a\left(b^2+c^2\right)\);\(y=b\left(a^2+c^2\right)\);\(z=c\left(b^2+a^2\right)\)

VT=\(\sqrt{\frac{x+y}{z}}+\sqrt{\frac{y+z}{x}}+\sqrt{\frac{x+z}{y}}\ge3\sqrt[6]{\frac{\left(x+y\right)\left(y+z\right)\left(z+x\right)}{xyz}}\ge3\sqrt{2}\)(BĐT Cô-si)

Dấu''='' xra\(\Leftrightarrow\)a=b=c

ĐK:x\(\ge-1\)(*)

bpt\(\Leftrightarrow3\left(x^2-x+1\right)+2\left(x+1\right)< 5\sqrt{\left(x+1\right)\left(x^2-x+1\right)}\)

\(\Leftrightarrow\left(\sqrt{x^2-x+1}-\sqrt{x+1}\right)\left(3\sqrt{x^2-x+1}-2\sqrt{x+1}\right)< 0\)

Đến đây bn chia 2 TH rồi giải bình thường nhá:D

Áp dụng BĐT Bunhiacopxki:

\(\sqrt{\left(a+b\right)\left(a+c\right)}\ge\sqrt{ac}+\sqrt{ab}\)

\(\Rightarrow\)\(\frac{a}{a+\sqrt{\left(a+b\right)\left(a+c\right)}}\)\(\le\frac{a}{a+\sqrt{ab}+\sqrt{ac}}\)=\(\frac{\sqrt{a}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}\)(1)

Tương tự ta có: \(\frac{b}{b+\sqrt{\left(b+c\right)\left(b+a\right)}}\le\frac{\sqrt{b}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}\)(2)

\(\frac{c}{c+\sqrt{\left(c+a\right)\left(c+b\right)}}\le\frac{\sqrt{c}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}\)(3)

Cộng theo vế của (1);(2)&(3) ta đc:

A\(\le1\)

Dấu''='' xảy ra\(\Leftrightarrow\)a=b=c

Use BĐT C-S ta có

x(1-yz)+y+z\(\le\sqrt{\left(x^2+\left(y+z\right)^2\right)\left(\left(1-yz\right)^2+1^2\right)}\)=\(\sqrt{\left(2+2yz\right)\left(2+\left(yz\right)^2-2yz\right)}\)

Vậy chỉ cần CM:\(\sqrt{\left(2+2yz\right)\left(2+\left(yz\right)^2-2yz\right)}\le2\)

\(\Leftrightarrow\left(1+yz\right)\left(2+\left(yz\right)^2-2yz\right)\le2\)

\(\Leftrightarrow\left(yz\right)^3\)\(\le\left(yz\right)^2\)

BĐT cuối cùng đúng vì:

2=x\(^2\)+y\(^2\)+z\(^2\)\(\ge\)y\(^2\)+z\(^2\)\(\ge\)2\(\left|yz\right|\)\(\Rightarrow\left|yz\right|\le1\)

\(\Rightarrow\left(yz\right)^3\)\(\le\)(yz)\(^2\)

BĐT đc chứng minh

đẳng thức xảy ra chẳng hạn 1 số =0 và 2 số =1

Ta có:P=(\(\frac{3a}{b+c}\)\(\frac{3a}{b+c}\)+3)+(\(\frac{4b}{a+c}\)+4)+(\(\frac{5c}{a+b}\)+5)-12

P=(a+b+c)(\(\frac{3}{b+c}\)+\(\frac{4}{c+a}\)+\(\frac{5}{a+b}\))-12

Áp dụng BĐT Bunhiacopxki

P=\(\frac{1}{2}\)((b+c)+(c+a)+(a+b))(\(\frac{3}{b+c}\)+\(\frac{4}{c+a}\)+\(\frac{5}{a+b}\))-12\(\ge\)\(\frac{\left(\sqrt{3}+2+\sqrt{5}\right)^2}{2}\)-12

Dấu''='' xảy ra \(\Leftrightarrow\)\(\frac{b+c}{\sqrt{3}}\)=\(\frac{c+a}{2}\)=\(\frac{a+b}{\sqrt{5}}\)

Ta có:\(\Delta\)l=4cm;A=8cm;T=2\(\pi\)\(\sqrt{\frac{\Delta l}{g}}\)=0,4(s)

2\(\alpha\)=\(\omega\)\(\Delta\)t nén

\(\Rightarrow\)\(\Delta\)t nén =\(\frac{2\alpha}{\omega}\)=\(\frac{2arccos\frac{\Delta l}{A}}{\frac{2\pi}{T}}\)=\(\frac{2.\frac{\pi}{3}}{2\pi}\).o,4=\(\frac{2}{15}\)(s)

Có j sai sót mong mn giúp đỡ