Ta biến đổi 1 tí nhé

\(\frac{4}{a}+\frac{5}{b}+\frac{3}{c}\ge4\left(\frac{3}{a+b}+\frac{2}{b+c}+\frac{1}{c+a}\right)\)

\(\Leftrightarrow\frac{3}{a+b}+\frac{2}{b+c}+\frac{1}{a+c}\le\frac{1}{4}\left(\frac{4}{a}+\frac{5}{b}+\frac{3}{c}\right)\)

Tới đây dễ dàng áp dụng BĐT \(\frac{4}{x+y}\le\frac{1}{x}+\frac{1}{y}\)

\(\Leftrightarrow\frac{3}{a+b}\le\frac{3}{4}.\frac{1}{a}+\frac{3}{4}.\frac{1}{b}\left(1\right)\)

\(\Leftrightarrow\frac{2}{b+c}\le\frac{1}{2}.\frac{1}{b}+\frac{1}{2}.\frac{1}{c}\left(2\right)\)

\(\Leftrightarrow\frac{1}{a+c}\le\frac{1}{4}.\frac{1}{a}+\frac{1}{4}.\frac{1}{c}\left(3\right)\)

Cộng vế với vế của (1), (2), (3) suy ra 

\(\frac{3}{a+b}+\frac{2}{b+c}+\frac{1}{a+c}\le\frac{3}{4}\cdot\frac{1}{a}+\frac{3}{4}\cdot\frac{1}{b}+\frac{1}{2}\cdot\frac{1}{b}+\frac{1}{2}\cdot\frac{1}{c}+\frac{1}{4}\cdot\frac{1}{a}+\frac{1}{4}\cdot\frac{1}{c}\)

\(\Leftrightarrow\frac{3}{a+b}+\frac{2}{b+c}+\frac{1}{a+c}\le\frac{1}{a}+\frac{5}{4}\cdot\frac{1}{b}+\frac{3}{4}\cdot\frac{1}{b}\)

\(\Leftrightarrow\frac{3}{a+b}+\frac{2}{b+c}+\frac{1}{a+c}\le\frac{1}{4}\left(\frac{4}{a}+\frac{5}{b}+\frac{3}{c}\right)\)

\(\Leftrightarrow Dpcm\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có:

\(\frac{ab+ac}{2}=\frac{ba+bc}{3}=\frac{ca+cb}{4}=\frac{\left(ab+ac\right)+\left(ba+bc\right)-\left(ca+cb\right)}{2+3-4}=\frac{2ab}{1}\)

Tương tự \(\frac{ab+ac}{2}=\frac{bc+ba}{3}=\frac{ca+cb}{4}=\frac{2bc}{5}\)

\(\frac{ab+ac}{2}=\frac{ba+bc}{3}=\frac{ca+cb}{4}=\frac{2ac}{3}\)

Do đó \(\frac{2ab}{1}=\frac{2bc}{5}\Rightarrow\frac{a}{1}=\frac{c}{5}\Rightarrow\frac{a}{3}=\frac{c}{15}\)

\(\frac{2bc}{5}=\frac{2ac}{3}\Rightarrow\frac{b}{5}=\frac{a}{3}\)

Do vậy \(\frac{a}{3}=\frac{b}{5}=\frac{c}{15}\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có:

$\frac{ab+ac}{2}=\frac{ba+bc}{3}=\frac{ca+cb}{4}=\frac{\left(ab+ac\right)+\left(ba+bc\right)-\left(ca+cb\right)}{2+3-4}=\frac{2ab}{1}$

Tương tự $\frac{ab+ac}{2}=\frac{bc+ba}{3}=\frac{ca+cb}{4}=\frac{2bc}{5}$

$\frac{ab+ac}{2}=\frac{ba+bc}{3}=\frac{ca+cb}{4}=\frac{2ac}{3}$

Do đó $\frac{2ab}{1}=\frac{2bc}{5}\Rightarrow \frac{a}{1}=\frac{c}{5}\Rightarrow \frac{a}{3}=\frac{c}{15}$

$\frac{2bc}{5}=\frac{2ac}{3}\Rightarrow \frac{b}{5}=\frac{a}{3}$

Do vậy $\frac{a}{3}=\frac{b}{5}=\frac{c}{15}$

nhầm sorry đừng để ý

a+b+c=0 <=> (a+b+c)²=0

<=>a²+b²+c²+2(ab+bc+ca)=0

<=>a²+b²+c²=-2(ab+bc+ca)

<=>(a²+b²+c²)²=[-2(ab+bc+ca)]²

<=>a⁴+b⁴+c⁴+2(a²b²+b²c²+c²a²)=4(a²b²+b²c²+c²a²)

<=>a⁴+b⁴+c⁴=2(a²b²+b²c²+c²a²) (1)

Lại có  (ab+bc+ca)² = a²b²+b²c²+c²a²+2abc(a+b+c) = a²b²+b²c²+c²a² (vì a+b+c=0) (2)

Từ (1) và (2) => đpcm

<=> 4C = 4.( 1 + 4 + 4² + 4³ + .... + 4¹⁰⁰ )

<=> 4C = 4 + 4² + 4³ + 4⁴ + ..... + 4¹⁰¹

<=> 4C -C = ( 4 + 4² + 4³ + 4⁴ + ..... + 4¹⁰¹ ) - ( 1 + 4 + 4² + 4³ + .... + 4¹⁰⁰ )

<=> 3C = 4¹⁰¹ - 1

=> C = ( 4¹⁰¹ - 1 ) : 3

B : 3 = 4¹⁰¹ : 3

Vì ( 4¹⁰¹ - 1 ) : 3 < 4¹⁰¹ : 3 => C < B : 3

Vậy C < B : 3

4c=4+4^2+4^3+..+4^101

=>4c-c=(4+4^2+4^3+...+4^101)-(1+4+4^2+..+4^100)

=>3c=4^101-1

=>c=(4^101-1)/3

 Mà b=4^101=>b/3=4^101/3

 Ta thấy c=(4^101-1)/3<b/3=4^101/3

=>c<b/3(đpcm)

 Tick đi

kho qua