Question

Cho a,b,c là độ dài 3 cạnh của một tam giác chứng minh :

ab + bc + ca <= a² +b² +c²<= 2(ab+bc+ca)

Answer 1

ta có: \(a^2\)+\(b^2\)+\(c^2\)\(\ge\)ab+bc+ca

<=> \(a^2\)+\(b^2\)+\(c^2\)-ab-bc-ca\(\ge\)0

<=>2\(a^2\)+2\(b^2\)+2\(c^2\)-2ab-2bc-2ca\(\ge\)0

<=> (\(a^2\)-2ab+\(b^2\))+(\(b^2\)-2bc+\(c^2\))+(\(c^2\)-2ca+\(a^2\))\(\ge\)0

<=> \(\left(a-b\right)^2\)+\(\left(b-c\right)^2\)+\(\left(c-a\right)^2\)\(\ge\)0 (luôn đúng)

dấu = xảy ra khi a =b=c

 

Answer 2

a&#x2212;b&lt;c&lt;=&gt;a2+b2&#x2212;2ab&lt;c2" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:16.38px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal; word-wrap:normal" class="MathJax_CHTML mjx-chtml">a−b<c<=>a2+b2−2ab<c2

b&#x2212;c&lt;a&lt;=&gt;b2+c2&#x2212;2bc&lt;a2" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:16.38px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal; word-wrap:normal" class="MathJax_CHTML mjx-chtml">b−c<a<=>b2+c2−2bc<a2

a&#x2212;c&lt;b&lt;=&gt;a2+c2&#x2212;2ac&lt;b2" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:16.38px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal; word-wrap:normal" class="MathJax_CHTML mjx-chtml">a−c<b<=>a2+c2−2ac<b2

2(a2+b2+c2)&#x2212;2(ab+bc+ac)&lt;a2+b2+c2&lt;=&gt;2(ab+ac+bc)&gt;a2+b2+c2" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:16.38px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal; word-wrap:normal" class="MathJax_CHTML mjx-chtml">2(a2+b2+c2)−2(ab+bc+ac)<a2+b2+c2<=>2(ab+ac+bc)>a2+b2+c2 (đpcm)

 

Answer 3

Bài này khó lắm tớ mới làm có vế trái thôi

Answer 4

\(a^{^{ }2}\)+\(b^{^{ }2}\)+\(c^{^{ }2}\)\(\le\)2(ab+bc+ca)

Vì a;b;c là 3 cạnh của 1 tam giác nên theo bất đẳng thức tam giác ta có

a\(\le\) b+c => a.a \(\le\)a.(b+c)=> \(a^{^{ }2}\)\(\le\)ab+ac(1)

b\(\le\)a+c => b.b\(\le\)b(a+c)=> \(b^{^{ }2}\)\(\le\)ab+ bc(2)

c\(\le\)a+b=> c.c\(\le\)c.(a+b) => \(c^{^{ }2}\)\(\le\)ac+bc(3)

cộng vế với vế của (1); (2) và (3) ta có:

\(a^{^{ }2}\)+\(b^{^{ }2}\)+\(c^{^{ }2}\)\(\le\) ab+ ac+ab+bc+ac+bc

vậy \(a^{^{ }2}\)+\(b^{^{ }2}\)+\(c^{^{ }2}\)\(\le\)2.(ab +bc+ca)

Answer 5

vế trái nữa bạn ơi