Áp dụng bất đẳng thức Cosi với các số dương a,b,c ta có:

\(\dfrac{a^2}{b+c}+\dfrac{b+c}{4}\ge2\sqrt{\dfrac{a^2\left(b+c\right)}{4\left(b+c\right)}}=a\) (1)

CMTT, ta có: \(\dfrac{b^2}{c+a}+\dfrac{c+a}{4}\ge b\) (2)

\(\dfrac{c^2}{a+b}+\dfrac{a+b}{4}\ge c\) (3)

Từ (1),(2) và (3) suy ra:

\(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}+\dfrac{2\left(a+b+c\right)}{4}\ge a+b+c\)

\(\Leftrightarrow\)\(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\) \(\ge\dfrac{a+b+c}{2}\) = \(\dfrac{6}{2}=3\)

\(\Rightarrow\) A\(\ge3\)

Dấu "=" xảy ra \(\Leftrightarrow\) \(a=b=c=2\)

Vậy GTNN của A = 3 \(\Leftrightarrow a=b=c=2\)

Lời giải:

Ta có: \(A=\frac{a+1}{a}+\frac{b+1}{b}+\frac{c+4}{c}\)

\(\Leftrightarrow A=1+\frac{1}{a}+1+\frac{1}{b}+1+\frac{4}{c}=3+\left(\frac{1}{a}+\frac{1}{b}+\frac{4}{c}\right)\)

Áp dụng BĐT Bunhiacopxky:

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

\(\Leftrightarrow \left(\frac{1}{a}+\frac{1}{b}+\frac{4}{c}\right)\geq \frac{4^2}{a+b+c}=\frac{16}{6}=\frac{8}{3}\)

Do đó: \(A\geq 3+\frac{8}{3}=\frac{17}{3}\) hay \(A_{\min}=\frac{17}{3}\)

Dấu bằng xảy ra khi \((a,b,c)=(\frac{3}{2}; \frac{3}{2}; 3)\)

Áp dụng Côsi:

\(a^2+\left(\frac{19-\sqrt{37}}{12}\right)^2\ge2\sqrt{\left(\frac{19-\sqrt{37}}{12}\right)^2.a^2}=2.\frac{19-\sqrt{37}}{12}a\)

\(b^2+\left(\frac{19-\sqrt{37}}{12}\right)^2\ge2.\frac{19-\sqrt{37}}{12}b\)

\(c^3+\left(\frac{\sqrt{37}-1}{6}\right)^3+\left(\frac{\sqrt{37}-1}{6}\right)^3\ge3\sqrt[3]{\left(\frac{\sqrt{37}-1}{6}\right)^3\left(\frac{\sqrt{37}-1}{6}\right)^3.c^3}=3.\left(\frac{\sqrt{37}-1}{6}\right)^2c\)

\(\Rightarrow a^2+b^2+c^3+2\left(\frac{19-\sqrt{37}}{12}\right)^2+2\left(\frac{\sqrt{37}-1}{6}\right)^3\ge2.\frac{19-\sqrt{37}}{12}a+2.\frac{19-\sqrt{37}}{12}b+3.\left(\frac{\sqrt{37}-1}{6}\right)^2c\)

\(\Rightarrow a^2+b^2+c^3+2.\left(\frac{19-\sqrt{37}}{12}\right)^2+3.\left(\frac{\sqrt{37}-1}{6}\right)^3\ge\frac{19-\sqrt{37}}{6}\left(a+b+c\right)=\frac{19-\sqrt{37}}{2}\)

\(\Rightarrow a^2+b^2+c^3\ge\frac{19-\sqrt{37}}{2}-2.\left(\frac{19-\sqrt{37}}{12}\right)^2-2.\left(\frac{\sqrt{37}-1}{6}\right)^3\)

Dấu "=" xảy ra khi và chỉ khi \(a=b=\frac{19-\sqrt{37}}{12};\text{ }c=\frac{\sqrt{37}-1}{6}\)

Vậy GTNN của biệu thức là .......

\(\left(a+b+c\right)^2=3a^2+3b^2+3c^2\)

\(\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ca=3a^2+3b^2+3c^2\)

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}a-b=0\\b-c=0\\c-a=0\end{matrix}\right.\) \(\Leftrightarrow a=b=c\)

\(\Rightarrow P=a^2+\left(a+2\right)\left(a+a\right)+2020\)

\(\Rightarrow P=3a^2+4a+2020=3\left(a+\frac{2}{3}\right)^2+\frac{6056}{3}\ge\frac{6056}{3}\)

\(P_{min}=\frac{6056}{3}\) khi \(a=-\frac{2}{3}\)

a) Áp dụng bđt AM-GM: \(+\hept{\begin{cases}x^2+y^2\ge2xy\\y^2+z^2\ge2yz\\z^2+x^2\ge2zx\end{cases}}\)\(\Rightarrow2\left(x^2+y^2+z^2\right)\ge2\left(xy+yz+zx\right)\)

\(\Leftrightarrow x^2+y^2+z^2\ge xy+yz+zx\left(đpcm\right)\)

Dấu "=" xay ra khi \(x=y=z\)

b) Bổ đề; \(x^2+y^2+z^2\ge\frac{\left(x+y+z\right)^2}{3}\)

Áp dụng : \(A=x^2+y^2+z^2\ge\frac{3^2}{3}=3\). Dấu "=" xảy ra khi \(x=y=z=1\)

c) Bổ đề: \(xy+yz+zx\le\frac{\left(x+y+z\right)^2}{3}\)

Áp dụng: \(B\le\frac{3^2}{3}=3\). Dấu "=" xảy ra khi \(x=y=z=1\)

d) \(A+B=x^2+y^2+z^2+xy+yz+zx=\left(x+y+z\right)^2-\left(xy+yz+zx\right)\)

\(\ge\left(x+y+z\right)^2-\frac{\left(x+y+z\right)^2}{3}\)

\(=\frac{2}{3}\left(x+y+z\right)^2=6\)

Dấu "=" xảy ra khi \(x=y=z=1\)

Bài này tuy dễ nhưng hơi loằng ngoằng giữa các câu :))

a. Cách phổ thông : x² + y² + z²\(\ge\)xy + yz + zx

<=> 2 ( x² + y² + z² )\(\ge\)2 ( xy + yz + zx )

<=> ( x² - 2xy + y² ) + ( y² - 2yz + z² ) + ( z² - 2zx + x² )\(\ge\)0

<=> ( x - y )² + ( y - z )² + ( z - x )²\(\ge\)0 ( * )

Vì ( x - y )² \(\ge\)0 ; ( y - z )² \(\ge\)0 ; ( z - x )²\(\ge\)0\(\forall\)x ; y ; z

=> ( * ) đúng 

=> A\(\ge\)B ; dấu "=" xảy ra <=> x = y = z

b. Xài Cauchy cho mới

( x² + y² + z² ) ( 1² + 1² + 1² )\(\ge\)( x + y + z )² = 3² = 9

<=> 3 ( x² + y² + z² )\(\ge\)9 

<=> x² + y² + z²\(\ge\)3

Dấu "=" xảy ra <=> x = y = z = 1

Vậy minA = 3 <=> x = y = z = 1

c. Theo câu a và câu b ta có : 3 ( xy + yz + zx )\(\le\)( x + y + z )² = 3² = 9

<=> xy + yz + zx\(\le\)3

Dấu "=" xảy ra <=> x = y = 1

Vậy maxB = 3 <=> x = y = 1

d. x + y + z = 3 . BP 2 vế ta được

x² + y² + z² + 2( xy + yz + zx ) = 9

Hay A + 2B = 9 . Mà B\(\le\)3 ( câu b )

=> A + B \(\ge\)6

Dấu "=" xảy ra <=> x = y = z = 1

Vậy min A + B = 6 <=> x = y = z = 1

b) Cái này là bạn đang chứng minh dùng CBS mà ?

Theo BĐT AM-GM :

\(M=a^2+b^2+c^2+d^2+ab+ac+bc+bd+dc+da\)

\(\ge10\sqrt[10]{\left(abcd\right)^5}=10\)

Dấu "=" xảy ra \(\Leftrightarrow a=b=c=d=1\)