1.

Ta sẽ chứng minh BĐT sau: \(\dfrac{1}{a^2+b^2}+\dfrac{1}{b^2+c^2}+\dfrac{1}{c^2+a^2}\ge\dfrac{10}{\left(a+b+c\right)^2}\)

Do vai trò a;b;c như nhau, ko mất tính tổng quát, giả sử \(c=min\left\{a;b;c\right\}\)

Đặt \(\left\{{}\begin{matrix}x=a+\dfrac{c}{2}\\y=b+\dfrac{c}{2}\end{matrix}\right.\) \(\Rightarrow x+y=a+b+c\)

Đồng thời \(b^2+c^2=\left(b+\dfrac{c}{2}\right)^2+\dfrac{c\left(3c-4b\right)}{4}\le\left(b+\dfrac{c}{2}\right)^2=y^2\)

Tương tự: \(a^2+c^2\le x^2\) ; \(a^2+b^2\le x^2+y^2\)

Do đó: \(A\ge\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{x^2+y^2}\)

Nên ta chỉ cần chứng minh: \(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{x^2+y^2}\ge\dfrac{10}{\left(x+y\right)^2}\)

Mà \(\dfrac{1}{\left(x+y\right)^2}\le\dfrac{1}{4xy}\) nên ta chỉ cần chứng minh:

\(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{x^2+y^2}\ge\dfrac{5}{2xy}\)

\(\Leftrightarrow\dfrac{1}{x^2}+\dfrac{1}{y^2}-\dfrac{2}{xy}+\dfrac{1}{x^2+y^2}-\dfrac{1}{2xy}\ge0\)

\(\Leftrightarrow\dfrac{\left(x-y\right)^2}{x^2y^2}-\dfrac{\left(x-y\right)^2}{2xy\left(x^2+y^2\right)}\ge0\)

\(\Leftrightarrow\dfrac{\left(x-y\right)^2\left(2x^2+2y^2-xy\right)}{2x^2y^2}\ge0\) (luôn đúng)

Vậy \(A\ge\dfrac{10}{\left(a+b+c\right)^2}\ge\dfrac{10}{3^2}=\dfrac{10}{9}\)

Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(\dfrac{3}{2};\dfrac{3}{2};0\right)\) và các hoán vị của chúng

2.

Ta có: \(B=\dfrac{ab+1-1}{1+ab}+\dfrac{bc+1-1}{1+bc}+\dfrac{ca+1-1}{1+ca}\)

\(B=3-\left(\dfrac{1}{1+ab}+\dfrac{1}{1+ca}+\dfrac{1}{1+ab}\right)\)

Đặt \(C=\dfrac{1}{1+ab}+\dfrac{1}{1+bc}+\dfrac{1}{1+ca}\)

Ta có: \(C\ge\dfrac{9}{3+ab+bc+ca}\ge\dfrac{9}{3+\dfrac{1}{3}\left(a+b+c\right)^2}=\dfrac{27}{13}\)

\(\Rightarrow B\le3-\dfrac{27}{13}=\dfrac{12}{13}\)

\(B_{max}=\dfrac{12}{13}\) khi \(a=b=c=\dfrac{2}{3}\)

Do \(a;b;c\in\left[0;1\right]\)

\(\Rightarrow\left(a-1\right)\left(b-1\right)\ge0\)\(\Leftrightarrow ab+1\ge a+b\)

\(\Leftrightarrow ab+c+1\ge a+b+c=2\)

\(\Rightarrow abc+ab+c+1\ge ab+c+1\ge2\)

\(\Rightarrow\left(c+1\right)\left(ab+1\right)\ge2\)

\(\Rightarrow\dfrac{1}{ab+1}\le\dfrac{c+1}{2}\)

Hoàn toàn tương tự, ta có: 

\(\dfrac{1}{bc+1}\le\dfrac{a+1}{2}\) ; \(\dfrac{1}{ca+1}\le\dfrac{b+1}{2}\)

Cộng vế: \(C\le\dfrac{a+b+c+3}{2}=\dfrac{5}{2}\)

\(\Rightarrow B\ge3-\dfrac{5}{2}=\dfrac{1}{2}\)

\(B_{min}=\dfrac{1}{2}\) khi \(\left(a;b;c\right)=\left(0;1;1\right)\) và các hoán vị của chúng

\(P=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}=\frac{9}{3}=3\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}\frac{1}{a}=\frac{1}{b}=\frac{1}{c}\\a+b+c=3\end{cases}}\Leftrightarrow a=b=c=1\)

Vậy.....

Chết,nhìn không kĩ đề. :(

ai giúp mình với

Lời giải:

Áp dụng BĐT Bunhiacopxky:

$(\frac{1}{a^4}+\frac{1}{b^4}+\frac{1}{c^4})(1+1+1)\geq (\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2})^2(1)$

$(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2})(1+1+1)\geq (\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^2(2)$

$(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})(a+b+c)\geq (1+1+1)^2$

$\Rightarrow \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq \frac{9}{a+b+c}=9$(3)$

Từ $(1); (2); (3)$ suy ra:
$\frac{1}{a^4}+\frac{1}{b^4}+\frac{1}{c^4}\geq \frac{9^4}{27}=243$
Vậy GTNN của biểu thức là 243 khi $a=b=c=\frac{1}{3}$

Đặt \(P=\dfrac{1}{a^4}+\dfrac{1}{b^4}+\dfrac{1}{c^4}=\left(\dfrac{1}{a^4}+\dfrac{1}{b^4}+\dfrac{1}{c^4}\right)\left(a+b+c\right)^4\) (do \(a+b+c=1\))

\(P=\left(\dfrac{1}{a^4}+\dfrac{1}{b^4}+\dfrac{1}{c^4}\right)\left(a+b+c\right)^4\ge3\sqrt[3]{\dfrac{1}{a^4.b^4.c^4}}.\left(3\sqrt[3]{abc}\right)^4=3^5=243\)

\(P_{min}=243\) khi \(a=b=c=\dfrac{1}{3}\)

\(Q\le\sqrt{3\left(a+b+b+c+c+a\right)}=\sqrt{6\left(a+b+c\right)}\le\sqrt{6.\sqrt{3\left(a^2+b^2+c^2\right)}}=\sqrt{6\sqrt{3}}\)

Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{\sqrt{3}}\)

Lại có:

\(a^2+b^2+c^2\le1\Rightarrow0\le a;b;c\le1\)

\(\Leftrightarrow a\left(a-1\right)+b\left(b-1\right)+c\left(c-1\right)\le0\)

\(\Leftrightarrow a+b+c\ge a^2+b^2+c^2=1\)

Do đó:

\(Q^2=2\left(a+b+c\right)+2\sqrt{a^2+ab+bc+ca}+2\sqrt{b^2+ab+bc+ca}+2\sqrt{c^2+ab+bc+ca}\)

\(Q^2\ge2\left(a+b+c\right)+2\sqrt{a^2}+2\sqrt{b^2}+2\sqrt{c^2}\)

\(Q^2\ge4\left(a+b+c\right)\ge4\)

\(\Rightarrow Q\ge2\)

Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(0;0;1\right)\) và hoán vị

\(P\le a^2+b^2+c^2+3\sqrt{3\left(a^2+b^2+c^2\right)}=12\)

\(P_{max}=12\) khi \(a=b=c=1\)

Lại có: \(\left(a+b+c\right)^2=3+2\left(ab+bc+ca\right)\ge3\Rightarrow a+b+c\ge\sqrt{3}\)

\(a+b+c\le\sqrt{3\left(a^2+b^2+c^2\right)}=3\)

\(\Rightarrow\sqrt{3}\le a+b+c\le3\)

\(P=\dfrac{\left(a+b+c\right)^2-\left(a^2+b^2+c^2\right)}{2}+3\left(a+b+c\right)\)

\(P=\dfrac{1}{2}\left(a+b+c\right)^2+3\left(a+b+c\right)-\dfrac{3}{2}\)

Đặt \(a+b+c=x\Rightarrow\sqrt{3}\le x\le3\)

\(P=\dfrac{1}{2}x^2+3x-\dfrac{3}{2}=\dfrac{1}{2}\left(x-\sqrt{3}\right)\left(x+6+\sqrt{3}\right)+3\sqrt{3}\ge3\sqrt{3}\)

\(P_{min}=3\sqrt{3}\) khi \(x=\sqrt{3}\) hay \(\left(a;b;c\right)=\left(0;0;\sqrt{3}\right)\) và hoán vị

thế bạn bt hok

Áp dụng bất đẳng thức Cauchy cho 2 số dương ta có:

a 2 + b 2 ≥ 2 a b ,   b 2 + c 2 ≥ 2 b c ,   c 2 + a 2 ≥ 2 c a  

Do đó:  2 a 2 + b 2 + c 2 ≥ 2 ( a b + b c + c a ) = 2.9 = 18 ⇒ 2 P ≥ 18 ⇒ P ≥ 9

Dấu bằng xảy ra khi  a = b = c = 3 . Vậy MinP= 9 khi  a = b = c = 3

Vì  a ,   b ,   c   ≥ 1 , nên  ( a − 1 ) ( b − 1 ) ≥ 0 ⇔ a b − a − b + 1 ≥ 0 ⇔ a b + 1 ≥ a + b

Tương tự ta có  b c + 1 ≥ b + c ,   c a + 1 ≥ c + a  

Do đó  a b + b c + c a + 3 ≥ 2 ( a + b + c ) ⇔ a + b + c ≤ 9 + 3 2 = 6

Mà   P = a 2 + b 2 + c 2 = a + b + c 2 − 2 a b + b c + c a = a + b + c 2 – 18

⇒ P ≤ 36 − 18 = 18 . Dấu bằng xảy ra khi :  a = 4 ; b = c = 1 b = 4 ; a = c = 1 c = 4 ; a = b = 1

Vậy maxP= 18 khi :  a = 4 ; b = c = 1 b = 4 ; a = c = 1 c = 4 ; a = b = 1