dung roi day =4

hình như là 4 hay sao ý

\(a,b>0\)  hay  \(a.b>0\)  vậy bạn? 

Mn ơi ai bt làm câu nào thì giúp mik cậu đó với !!

1. a. 

Ta có: 12⁸ = (12⁴)² = 20736²

Ta thấy: 20736 > 81

=> 12⁸ > 81²

(Các câu khác cũng tương tự nhé.)

\(P=\dfrac{9}{ab+bc+ca}+\dfrac{2}{a^2+b^2+c^2}\)

\(=2\left[\dfrac{1}{a^2+b^2+c^2}+\dfrac{4}{2\left(ab+bc+ca\right)}\right]+\dfrac{5}{ab+bc+ca}\)

\(\ge2.\dfrac{\left(1+2\right)^2}{\left(a+b+c\right)^2}+\dfrac{5}{ab+bc+ca}\)

\(=\dfrac{18}{1}+\dfrac{5}{ab+bc+ca}\ge18+5.\dfrac{3}{\left(a+b+c\right)^2}=18+15=33\)

Đẳng thức xảy ra khi a=b=c=1/3.

Vậy GTNN của P là 33.

Lời giải:
Áp dụng BĐT Cauchy-Schwarz:
\(P=2(\frac{1}{ab+bc+ac}+\frac{1}{ab+bc+ac}+\frac{1}{a^2+b^2+c^2})+\frac{1}{2(ab+bc+ac)}\\ \geq 2.\frac{9}{2(ab+bc+ac)+a^2+b^2+c^2}+\frac{1}{2(ab+bc+ac)}\\ =\frac{18}{(a+b+c)^2}+\frac{1}{2(ab+bc+ac)}\\ =18+\frac{1}{2(ab+bc+ac)}\)

Áp dụng BĐT AM-GM:

$2(ab+bc+ac)\leq 2.\frac{(a+b+c)^2}{3}=\frac{2}{3}$

$\Rightarrow \frac{1}{2(ab+bc+ac)}\geq \frac{3}{2}$

$\Rightarrow P\geq 18+\frac{3}{2}=\frac{39}{2}$
Vậậy $P_{\min}=\frac{39}{2}$ khi $a=b=c=\frac{1}{3}$

áp dụng bất đẳng thức phụ \(\dfrac{1}{a}+\dfrac{1}{b}\)≥\(\dfrac{4}{a+b}\)<=>(a-b)²≥0 (luôn đúng)
Ta có P≥\(\dfrac{\left(3+\sqrt{2}\right)^2}{\left(a+b+c\right)^2}\)=(3+\(\sqrt{2}\))²
Dấu = xảy ra <=> a=b=c=1/3

\(M=\dfrac{1}{a^2+b^2+c^2}+\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\ge\dfrac{1}{a^2+b^2+c^2}+\dfrac{9}{ab+bc+ca}=\dfrac{1}{a^2+b^2+c^2}+\dfrac{1}{ab+bc+ca}+\dfrac{1}{ab+bc+ca}+\dfrac{1}{ab+bc+ca}+\dfrac{7}{ab+bc+ca}\)

\(M\ge\dfrac{9}{a^2+b^2+c^2+2ab+2bc+2ca}+\dfrac{7}{ab+bc+ca}=\dfrac{9}{\left(a+b+c\right)^2}+\dfrac{7}{ab+bc+ca}\)

\(ab+bc+ca\le\dfrac{\left(a+b+c\right)^2}{3}\)

\(\Rightarrow M\ge\dfrac{9}{\left(a+b+c\right)^2}+\dfrac{7}{ab+bc+ca}=9+\dfrac{7.3}{\left(a+b+c\right)^2}=9+21=30\)

\(Min_M=30\Leftrightarrow a=b=c=\dfrac{1}{3}\)

Áp dụng BĐT Svacxo

\(m\text{≥}\dfrac{1}{a^2+b^2+c^2}+\dfrac{9}{ab+bc+ca}\)

\(=\dfrac{1}{a^2+b^2+c^2}+\dfrac{1}{ab+bc+ca}+\dfrac{1}{ab+bc+ca}+\dfrac{7}{ab+bc+ca}\)

≥ \(\dfrac{9}{a^2+b^2+c^2+2\left(ab+bc+ca\right)}\)\(+\dfrac{7}{ab+bc+ca}\)

\(=\dfrac{9}{\left(a+b+c\right)^2}+\dfrac{7}{ab+bc+ca}\)

CM BĐT: \(a^2+b^2+c^2\text{≥}ab+bc+ca\)

⇔ \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\text{≥}0\) (luôn đúng)

⇒ \(\left(a+b+c\right)^2\text{≥}3\left(ab+bc+ca\right)\)

⇒ \(\dfrac{\left(a+b+c\right)^2}{3}\text{≥}ab+bc+ca\)

⇒ \(m\text{≥}\dfrac{9}{\left(a+b+c\right)^2}+\dfrac{7}{\dfrac{\left(a+b+c\right)^2}{3}}=9+21=30\) 

(vì a+b+c=1)

Vậy...