Chủ đề này có 2 trả lời

[Thong Nhat]

a) Cho x, y, z > 0. Chứng minh: $\sqrt{xy(x+y)}+\sqrt{yz(y+z)}+\sqrt{zx(z+x)}\geq \sqrt{2xyz}+\sqrt{(x+y)(y+z)(z+x)}$

b) Cho a, b, c > 0 thỏa a+b+c=3. Chứng minh rằng: $\frac{5-3bc}{1+a}+\frac{5-3ca}{1+b}+\frac{5-3ab}{1+c}\geq ab+bc+ca$

[DOTOANNANG]

$\lceil\,\,\text{a}\,\,\rfloor$

Bất đẳng thức chặt hơn:

$\sqrt{xy\left ( x+ y \right )}+ \sqrt{yz\left ( y+ z \right )}+ \sqrt{zx\left ( z+ x \right )}$ $\geqq \sqrt{\left ( x+ y \right )\left ( y+ z \right )\left ( z+ x \right )}+ \left ( x+ y+ z \right )\sqrt{\frac{2\,xyz}{3\left ( xy+ yz+ zx \right )}}$

 

$\Leftrightarrow \sqrt{\frac{xy}{\left ( x+ z \right )\left ( y+ z \right )}}+ \sqrt{\frac{yz}{\left ( y+ x \right )\left ( z+ x \right )}}+ \sqrt{\frac{zx}{\left ( z+ y \right )\left ( x+ y \right )}}$ $\geqq 1+ \left ( x+ y+ z \right )\sqrt{\frac{2\,xyz}{3\left ( xy+ yz+ zx \right )\left ( x+ y \right )\left ( y+ z \right )\left ( z+ x \right )}}$

 

Đặt: $\{\begin{array}{ll}a= \sqrt{\frac{yz}{\left ( y+ x \right )\left ( z+ x \right )}}\\ \\ b= \sqrt{\frac{zx}{\left ( z+ y \right )\left ( x+ y \right )}}\\ \\ c= \sqrt{\frac{xy}{\left ( x+ z \right )\left ( y+ z \right )}} \\ \end{array}\,\,\rightarrow \,\,a^{2}+ b^{2}+ c^{2}+ 2\,abc= 1$

 

Do đó bất đẳng thức trên tương đương với:

 

$1+ \sqrt{2\left ( 1- \sqrt{\frac{xy}{\left ( x+ z \right )\left ( y+ z \right )}} \right )\left ( 1- \sqrt{\frac{yz}{\left ( y+ x \right )\left ( z+ x \right )}} \right )\left ( 1- \sqrt{\frac{zx}{\left ( z+ y \right )\left ( x+ y \right )}} \right )}$ $\geqq 1+ \left ( x+ y+ z \right )\sqrt{\frac{2\,xyz}{3\left ( xy+ yz+ zx \right )\left ( x+ y \right )\left ( y+ z \right )\left ( z+ x \right )}}$

 

hay:

 

$$\prod\limits_{cyc}\left ( \sqrt{\left ( x+ z \right )\left ( y+ z \right )}- \sqrt{xy} \right )\geqq \frac{xyz\left ( x+ y+ z \right )^{2}}{3\left ( xy+ yz+ zx \right )}$$

 

$\Leftrightarrow \prod\limits_{cyc} \left ( \sqrt{\left ( x+ z \right )\left ( y+ z \right )}+ \sqrt{xy} \right )\leqq 3\left ( x+ y+ z \right )\left ( xy+ yz+ zx \right )$

 

$$\Leftrightarrow \left ( 1+ a \right )\left ( 1+ b \right )\left ( 1+ c \right )\leqq 3\left ( 1+ abc \right )\Leftrightarrow \sum\limits_{cyc}a+ \sum\limits_{cyc}ab\leqq 2+ 2\,abc$$

 

[DOTOANNANG]

$\lceil\,\,\text{a}\,\,\rfloor$

Bất đẳng thức chặt hơn:

$\sqrt{xy\left ( x+ y \right )}+ \sqrt{yz\left ( y+ z \right )}+ \sqrt{zx\left ( z+ x \right )}$ $\geqq \sqrt{\left ( x+ y \right )\left ( y+ z \right )\left ( z+ x \right )}+ \left ( x+ y+ z \right )\sqrt{\frac{2\,xyz}{3\left ( xy+ yz+ zx \right )}}$

 

$\Leftrightarrow \sqrt{\frac{xy}{\left ( x+ z \right )\left ( y+ z \right )}}+ \sqrt{\frac{yz}{\left ( y+ x \right )\left ( z+ x \right )}}+ \sqrt{\frac{zx}{\left ( z+ y \right )\left ( x+ y \right )}}$ $\geqq 1+ \left ( x+ y+ z \right )\sqrt{\frac{2\,xyz}{3\left ( xy+ yz+ zx \right )\left ( x+ y \right )\left ( y+ z \right )\left ( z+ x \right )}}$

 

Đặt: $\{\begin{array}{ll}a= \sqrt{\frac{yz}{\left ( y+ x \right )\left ( z+ x \right )}}\\ \\ b= \sqrt{\frac{zx}{\left ( z+ y \right )\left ( x+ y \right )}}\\ \\ c= \sqrt{\frac{xy}{\left ( x+ z \right )\left ( y+ z \right )}} \\ \end{array}\,\,\rightarrow \,\,a^{2}+ b^{2}+ c^{2}+ 2\,abc= 1$

 

Do đó bất đẳng thức trên tương đương với:

 

$1+ \sqrt{2\left ( 1- \sqrt{\frac{xy}{\left ( x+ z \right )\left ( y+ z \right )}} \right )\left ( 1- \sqrt{\frac{yz}{\left ( y+ x \right )\left ( z+ x \right )}} \right )\left ( 1- \sqrt{\frac{zx}{\left ( z+ y \right )\left ( x+ y \right )}} \right )}$ $\geqq 1+ \left ( x+ y+ z \right )\sqrt{\frac{2\,xyz}{3\left ( xy+ yz+ zx \right )\left ( x+ y \right )\left ( y+ z \right )\left ( z+ x \right )}}$

 

hay:

 

$$\prod\limits_{cyc}\left ( \sqrt{\left ( x+ z \right )\left ( y+ z \right )}- \sqrt{xy} \right )\geqq \frac{xyz\left ( x+ y+ z \right )^{2}}{3\left ( xy+ yz+ zx \right )}$$

 

$\Leftrightarrow \prod\limits_{cyc} \left ( \sqrt{\left ( x+ z \right )\left ( y+ z \right )}+ \sqrt{xy} \right )\leqq 3\left ( x+ y+ z \right )\left ( xy+ yz+ zx \right )$

 

$$\Leftrightarrow \left ( 1+ a \right )\left ( 1+ b \right )\left ( 1+ c \right )\leqq 3\left ( 1+ abc \right )\Leftrightarrow \sum\limits_{cyc}a+ \sum\limits_{cyc}ab\leqq 2+ 2\,abc$$

Đặt: $\{\begin{array}{ll}a= \frac{u}{2}\\ \\ b= \frac{v}{2}\\ \\ c= \frac{w}{2}\end{array}\,\,\rightarrow \,\,u^{2}+ v^{2}+ w^{2}+ uvw= 4$

 

Bất đẳng thức cuối cùng ta cần chứng minh tương đương với:

 

$$2\left ( u+ v+ w \right )+ uv+ vw+ wu\leqq 8+ uvw$$

 

Với $u^{2}+ v^{2}+ w^{2}+ uvw= 4$ ta sẽ được: $uv+ vw+ wu- uvw\leqq 2$. ( $\lceil$ Titu Andreescu $\rfloor$ $-$ USAMO $2001$ )

https://nguyenhuyena...uat-usamo-2001/

 

Việc còn lại là chứng minh: $2\left ( u+ v+ w \right )\leqq 6$ hay $u+ v+ w\leqq 3$

https://diendantoanh...-4/#entry717544