\(\left(x+1\right)\left(y+1\right)=9\)

\(\Rightarrow xy+x+y+1=9\)

\(\Rightarrow xy+x+y=8\)

\(\Rightarrow x+y=8-xy\left(1\right)\)

\(K=x^2+y^2\)

\(\Rightarrow K=\left(x+y\right)^2-2xy=\left(8-xy\right)^2-2xy\)

\(\Rightarrow K=64-16xy+\left(xy\right)^2-2xy\)

\(\Rightarrow K=\left(xy\right)^2-18xy+64\)

\(\Rightarrow K=\left(xy\right)^2-18xy+81-17\)

\(\Rightarrow K=\left(xy-9\right)^2-17\ge-17\left(\left(xy-9\right)^2\ge0,\forall x;y\right)\)

\(\Rightarrow GTNN\left(K\right)=-17\)

GTNN (K) = -17

$\left(x+1\right)\left(y+1\right)=9$

$\Rightarrow xy+x+y+1=9$

$\Rightarrow xy+x+y=8$

$\Rightarrow x+y=8-xy\left(1\right)$

�=�2+�2K=x2+y2

⇒�=(�+�)2−2��=(8−��)2−2��⇒K=(x+y)2−2xy=(8−xy)2−2xy

⇒�=64−16��+(��)2−2��⇒K=64−16xy+(xy)2−2xy

⇒�=(��)2−18��+64⇒K=(xy)2−18xy+64

⇒�=(��)2−18��+81−17⇒K=(xy)2−18xy+81−17

⇒�=(��−9)2−17≥−17((��−9)2≥0,∀�;�)⇒K=(xy−9)2−17≥−17((xy−9)2≥0,∀x;y)

$\Rightarrow GTNN\left(K\right)=-17$

Ta có: \(\left(x-y\right)\left(1-xy\right)\le\dfrac{1}{4}\left(x-y+1-xy\right)^2=\dfrac{1}{4}\left(x+1\right)^2\left(1-y\right)^2\)

\(\Rightarrow P\le\dfrac{\left(1+x\right)^2\left(1-y\right)^2}{4\left(1+x\right)^2\left(1+y\right)^2}=\dfrac{1}{4}\left(\dfrac{y^2-2y+1}{y^2+2y+1}\right)=\dfrac{1}{4}\left(1-\dfrac{4y}{y^2+2y+1}\right)\le\dfrac{1}{4}\)

\(P_{max}=\dfrac{1}{4}\) khi \(\left(x;y\right)=\left(1;0\right)\)

Lại có:

\(\left(y-x\right)\left(1-xy\right)\le\dfrac{1}{4}\left(y-x+1-xy\right)^2=\dfrac{1}{4}\left(1+y\right)^2\left(1-x\right)^2\)

\(\Rightarrow-P\le\dfrac{\left(1+y\right)^2\left(1-x\right)^2}{4\left(1+y\right)^2\left(1+x\right)^2}=\dfrac{1}{4}\left(\dfrac{1-2x+x^2}{1+2x+x^2}\right)=\dfrac{1}{4}\left(1-\dfrac{4x}{x^2+2x+1}\right)\le\dfrac{1}{4}\)

\(\Rightarrow-P\le\dfrac{1}{4}\Rightarrow P\ge-\dfrac{1}{4}\)

\(P_{min}=-\dfrac{1}{4}\) khi \(\left(x;y\right)=\left(0;1\right)\)

(Do \(y\ge0\Rightarrow\dfrac{4y}{y^2+2y+1}\ge0\Rightarrow1-\dfrac{4y}{y^2+2y+1}\le1\Rightarrow...\))

1) Áp dụng bất đẳng thức AM - GM và bất đẳng thức Schwarz:

\(P=\dfrac{1}{a}+\dfrac{1}{\sqrt{ab}}\ge\dfrac{1}{a}+\dfrac{1}{\dfrac{a+b}{2}}\ge\dfrac{4}{a+\dfrac{a+b}{2}}=\dfrac{8}{3a+b}\ge8\).

Đẳng thức xảy ra khi a = b = \(\dfrac{1}{4}\).

2.

\(4=a^2+b^2\ge\dfrac{1}{2}\left(a+b\right)^2\Rightarrow a+b\le2\sqrt{2}\)

Đồng thời \(\left(a+b\right)^2\ge a^2+b^2\Rightarrow a+b\ge2\)

\(M\le\dfrac{\left(a+b\right)^2}{4\left(a+b+2\right)}=\dfrac{x^2}{4\left(x+2\right)}\) (với \(x=a+b\Rightarrow2\le x\le2\sqrt{2}\) )

\(M\le\dfrac{x^2}{4\left(x+2\right)}-\sqrt{2}+1+\sqrt{2}-1\)

\(M\le\dfrac{\left(2\sqrt{2}-x\right)\left(x+4-2\sqrt{2}\right)}{4\left(x+2\right)}+\sqrt{2}-1\le\sqrt{2}-1\)

Dấu "=" xảy ra khi \(x=2\sqrt{2}\) hay \(a=b=\sqrt{2}\)

3. Chia 2 vế giả thiết cho \(x^2y^2\)

\(\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{x^2}+\dfrac{1}{y^2}-\dfrac{1}{xy}\ge\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2\)

\(\Rightarrow0\le\dfrac{1}{x}+\dfrac{1}{y}\le4\)

\(A=\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}-\dfrac{1}{xy}\right)=\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2\le16\)

Dấu "=" xảy ra khi \(x=y=\dfrac{1}{2}\)

Cách 1:

Áp dụng tính chất cuẩ BĐT, Ta có: \(x^4+y^4+z^4\ge\frac{\left(x^2+y^2+z^2\right)^2}{3}\)

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

=> \(x^4+y^4+z^4\ge\frac{\left(\frac{x+y+z}{3}\right)^2}{3}=\frac{16}{27}\)

=> GTNN của \(x^4+y^4+z^4=\frac{16}{27}\) đạt được khi x=y=z=2/3

bạn còn cách 2 ko?

Ap dung bat dang thuc a²+b²+c²>=\(\frac{\left(a+b+c\right)^2}{3}\)

=> x⁴+y⁴+z⁴>=\(\frac{\left(x^2+y^2+z^2\right)^2}{3}>=\frac{\left(x+y+z\right)^4}{27}=\frac{16}{27}\)

 dau = xay ra khi x=y=z=2/3

a)
Cho x, y, z thỏa mãn x^2 + y^2 + z^2 ≤ 3. Tìm GTNN của P = 1/(1 + xy) + 1/(1 + yz) + 1/(1 + xz)
Engel
P≥(1+1+1)^2/[3+(xy+yz+zx)]
BDT co ban
x^2 + y^2 + z^2 ≥xy+yz+zx
=>xy+yz+zx≤3
<=>P≥3^2/(3+3)=3/2
gtnn P=3/2 "= khi x=y=z=1

Đề bài sai, phản ví dụ:

Với \(x=1;y=0\) thì x;y thỏa mãn \(\left(x+1\right)\left(y+1\right)=2\)

Nhưng \(P=1-\sqrt{6}\) không phải số nguyên

Đáp án đúng : B