\(3=x+y+xy\le\sqrt{2\left(x^2+y^2\right)}+\dfrac{x^2+y^2}{2}\)

\(\Rightarrow\left(\sqrt{x^2+y^2}-\sqrt{2}\right)\left(\sqrt{x^2+y^2}+3\sqrt{2}\right)\ge0\)

\(\Rightarrow x^2+y^2\ge2\)

\(\Rightarrow-\left(x^2+y^2\right)\le-2\)

\(P=\sqrt{9-x^2}+\sqrt{9-y^2}+\dfrac{x+y}{4}\le\sqrt{2\left(9-x^2+9-y^2\right)}+\dfrac{\sqrt{2\left(x^2+y^2\right)}}{4}\)

\(P\le\sqrt{2\left(18-x^2-y^2\right)}+\dfrac{1}{4}.\sqrt{2\left(x^2+y^2\right)}\)

\(P\le\left(\sqrt{2}-1\right)\sqrt{18-x^2-y^2}+\sqrt[]{2}\sqrt{\dfrac{\left(18-x^2-y^2\right)}{2}}+\dfrac{1}{2}\sqrt{\dfrac{x^2+y^2}{2}}\)

\(P\le\left(\sqrt{2}-1\right).\sqrt{18-2}+\sqrt{\left(2+\dfrac{1}{4}\right)\left(\dfrac{18-x^2-y^2+x^2+y^2}{2}\right)}=\dfrac{1+8\sqrt{2}}{2}\)

Dấu "=" xảy ra khi \(x=y=1\)

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

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

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

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

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

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

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

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

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

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

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

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)

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

\(2x^2+3y^2+4z^2=21\Rightarrow2x^2\le21-3.1^2-4.1^2=14\)

\(\Rightarrow x\le\sqrt{7}\)

Tương tự ta có \(y\le\sqrt{5}\) và \(z\le2\)

Do đó:

\(\left(z-1\right)\left(z-2\right)\le0\Rightarrow z^2+2\le3z\Rightarrow4z^2+8\le12z\) (1)

\(\left(x-1\right)\left(2x-10\right)\le0\Rightarrow2x^2+10\le12x\) (2)

\(\left(y-1\right)\left(3y-9\right)\le0\Leftrightarrow3y^2+9\le12y\) (3)

Cộng vế (1);(2) và (3):

\(\Rightarrow12\left(x+y+z\right)\ge2x^2+3y^2+4z^2+27\ge48\)

\(\Rightarrow x+y+z\ge4\)

\(M_{min}=4\) khi \(\left(x;y;z\right)=\left(1;1;2\right)\)

Theo chứng minh ban đầu ta có: \(z\le2\Rightarrow z-2\le0\)

Theo giả thiết \(z\ge1\Rightarrow z-1\ge0\)

\(\Rightarrow\left(z-1\right)\left(z-2\right)\le0\)

Tương tự: \(x< \sqrt{5}< 5\Rightarrow x-5< 0\Rightarrow2x-10< 0\)

\(\Rightarrow\left(x-1\right)\left(2x-10\right)\le0\)

y cũng như vậy

Lời giải;

Vế 1:

Áp dụng BĐT AM-GM:

$2=(x^2+y^2)(1+1)\geq (x+y)^2\Rightarrow x+y\leq \sqrt{2}$

$x^3+\frac{x}{2}\geq \sqrt{2}x^2$

$y^3+\frac{y}{2}\geq \sqrt{2}y^2$

$\Rightarrow x^3+y^3+\frac{x+y}{2}\geq \sqrt{2}(x^2+y^2)=\sqrt{2}$

$\Rightarrow x^3+y^3\geq \sqrt{2}-\frac{x+y}{2}\geq \sqrt{2}-\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}$

-----------------------

Vế 2:

$x^2+y^2=1$

$\Rightarrow x^2=1-y^2\leq 1\Rightarrow -1\leq x\leq 1$

$y^2=1-x^2\leq 1\Rightarrow -1\leq y\leq 1$

$\Rightarrow x^3\leq x^2; y^3\leq y^2$

$\Rightarrow x^3+y^3\leq x^2+y^2$ hay $x^3+y^3\leq 1$

\(\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\ge\frac{\left(x+y+x\right)^2}{y+z+z+x+x+y}=\frac{x+y+x}{2}=1\)

Dấu ' =' xảy ra khi \(x=y=z=\frac{2}{3}\)

Đáp án A

Sử dụng BĐT buhinhacopski ta có

x − 2 + y + 3 2 ≤ 1 + 1 x − 2 + y + 3 = 2 x + y + 2 .

Tức là ta có  x + y + 1 2 ≤ 4 2 x + y + 2   . Đặt  t = x + y   . Chú ý rằng  t ≥ − 1   .

Ta có

t + 1 2 ≤ 8 t + 8 ⇔ t 2 − 6 t − 7 ≤ 0 ⇔ − 1 ≤ t ≤ 7.  

Vậy max t = 7  xảy ra khi   x − 2 = y + 3 x + y = 7 ⇔ x = 6 y = 1 .