3 replies to this topic

[arsfanfc]

Cho $a,b,c>0$ CM $\sqrt{a^2-ab+b^2}+\sqrt{a^2-ac+c^2} \geq \sqrt{b^2+bc+c^2}$

[hoanglong2k]

Cho $a,b,c>0$ CM $\sqrt{a^2-ab+b^2}+\sqrt{a^2-ac+c^2} \geq \sqrt{b^2+bc+c^2}$

 

 

 Áp dụng BĐT Minkowski ta có :

$$\sqrt{a^2-ab+b^2}+\sqrt{a^2-ac+c^2}=\sqrt{\left ( \frac{b}{2}-a \right )^2+\frac{3b^2}{4}}+\sqrt{\left ( a-\frac{c}{2} \right )^2+\frac{3c^2}{4}}$$

$$\geq \sqrt{\frac{(b-c)^2}{4}+\frac{3(b+c)^2}{4}}=\sqrt{b^2+bc+c^2}$$

[HoangVienDuy]

Cho $a,b,c>0$ CM $\sqrt{a^2-ab+b^2}+\sqrt{a^2-ac+c^2} \geq \sqrt{b^2+bc+c^2}$

áp dụng Minkowsky ta có

$\sqrt{(a-\frac{1}{2}b)^{2}+\frac{3}{4}b^{2}}+\sqrt{(\frac{1}{2}c-a)^{2}+\frac{3}{4}c^{2}}$

$\geq \sqrt{(a-\frac{1}{2}b+\frac{1}{2}c-a)^{2}+\frac{3}{4}(b+c)^{2}}=\sqrt{\frac{1}{4}(c-b)^{2}+\frac{3}{4}(b+c)^{2}}=\sqrt{b^{2}+bc+c^{2}}$

Edited by Hoang Nhat Tuan, 13-07-2015 - 17:07.

[Quoc Tuan Qbdh]

Áp dụng BĐT $Min-cop-xki$ 

$\sqrt{a^{2}-ab+b^{2}}+\sqrt{a^{2}-ac+c^{2}}=\sqrt{(a-\frac{b}{2})^{2}+(\frac{\sqrt{3}b}{2})^{2}}+\sqrt{(\frac{c}{2}-a)^{2}+\left ( \frac{\sqrt{3}c}{2} \right )^{2}}\geq \sqrt{(\frac{c}{2}-\frac{b}{2})^{2}+\frac{3}{4}(b+c)^{2}}=\sqrt{b^{2}+bc+c^{2}}$